Answer:
18.015
Explanation:
If a hydrogen of an alkane is replaced by NH, the compound becomes_________
a. alcohol
b. carboxylic acid
c. phenol
d. amine
Answer:
d. amine.
It becomes an amine.
Explanation:
With general formular
[tex]{ \bf{primary \: amine :R - NH _{2}}} \\ { \bf{secondary \: amine : R {}^{i} - NH - R}} \\ { \bf{tertiary \: amine :R {}^{ii} - N(R {}^{i} ) - R }}[/tex]
R is the aryl group such as alkane
You have been contracted to determine how different salts affect the pH of water. Which of the solids in the following set should you test to investigate for the effects of cations on pH?
a. AlBr3
b. Rb2SO3
c. MgCl2
d. RbBrO
e. CH3NH3Br
Answer:
Hence the solids that should test to investigate the effects of cations on pH is
[tex]AlBr_{3}[/tex] (Cation is Al 3+)
[tex]MgCl_{2}[/tex] ( Cation is Mg 2+)
[tex]CH_{3} NH_{3} Br[/tex] ( Cation is NH2+).
Explanation:
The solids in the following should you test to investigate the effects of cations on pH.
[tex]AlBr_{3}[/tex] contains (Cation is Al 3+)
[tex]MgCl_{2}[/tex] contains ( Cation is Mg 2+)
[tex]CH_{3} NH_{3} Br[/tex] contains( Cation is NH2+ )
The atoms or the molecules containing the positive charge that gets attracted to the cathode are called cations. The compounds a. [tex]\rm AlBr_{3}[/tex], c. [tex]\rm MgCl_{2}[/tex] and e. [tex]\rm CH_{3}NH_{3}Br[/tex] should be investigated.
What are cations and pH?Cations are the positive charge containing molecules and atoms that have more protons in their nucleus than the number of electrons in their shells. They are formed when they lose one or more electrons to another atom.
The addition or release of the electrons of the cations and anions affects the pH system as absorption of the cation decreases the pH and absorption of the anions increases the pH.
Hence, [tex]\rm Al^{3+}[/tex], [tex]\rm Mg^{2+}[/tex] and [tex]\rm NH^{2+}[/tex] are the cation that should be investigated. The addition of the cations will reduce the pH of the reaction.
Therefore, absorption of the cation reduces the pH.
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An analytical chemist is titrating of a solution of hydrazoic acid with a solution of . The of hydrazoic acid is . Calculate the pH of the acid solution after the chemist has added of the solution to it.
Answer:
pH = 12.43
Explanation:
...is titrating 212.7 mL of a 0.6800 M solution of hydrazoic acid (HN3) with a 0.2900 M solution of KOH. The p Ka of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 571.6 mL of the KOH solution to it.
To solve this question we need to know that hidrazoic acid reacts with KOH as follows:
HN3 + KOH → KN3 + H2O
Moles KOH:
0.5716L * (0.2900mol /L) =0.1658 moles of KOH
Moles HN3:
0.2127L * (0.6800mol/L) = 0.1446 moles HN3
As the reaction is 1:1, the KOH is in excess. The moles in excess of KOH are:
0.1658 moles - 0.1446 moles =
0.0212 mol KOH
In 212.7mL + 571.6mL = 784.3mL = 0.7843L
The molarity of KOH = [OH-] is:
0.0212 mol KOH / 0.7843L = 0.027M = [OH-]
The pOH is defined as -log [OH-]
pOH = -log 0.027M
pOH = 1.57
pH = 14 - pOH
pH = 12.43
A natural element consists of two isotopes: Cl-35 and Cl-37. The composition of these two isotopes differs by:
Answer:
There are no options in this question, however, it can be answered based on general understanding
- The number of neutrons each isotope contain
Explanation:
Isotopes are atoms of an element with the same atomic number or number of protons but different mass number/atomic masses. Since isotopes have same proton numbers, they have similar chemical behavior or identity.
However, difference in atomic mass or mass number of the same atomic number indicates that the number of neutrons each isotope contain differs from one another. Hence, in two isotopes of chlorine given as follows: Cl-35 and Cl-37, the NUMBER OF NEUTRONS in each atom differentiates the two isotopes.
Cl-35 contains 18 neutrons while Cl-37 contains 20 neutrons.
How many mL of 0.200M KI would contain 0.0500 moles of KI?
Please explain and show work.
Answer:
250ml
Explanation:
call it V
V*0.2=0.05 (moles)
so V=0.05/0.2 = 0.25l = 250ml
We know
[tex]\boxed{\Large{\sf Molarity=\dfrac{No\:of\:moles\:of\:solute}{Volume\:of\:solution\:in\;\ell}}}[/tex]
[tex]\\ \Large\sf\longmapsto Volume\:of\:KI=\dfrac{0.05}{0.2}[/tex]
[tex]\\ \Large\sf\longmapsto Volume\:of\:KI=0.25L[/tex]
[tex]\\ \Large\sf\longmapsto Volume\:of\:KI=250mL[/tex]
Give the change in condition to go from a gas to a solid. Group of answer choices cool or increase pressure cool or reduce pressure increase heat or reduce pressure increase heat or increase pressure none of the above
Answer:
cool or increase pressure
Explanation:
For a gas to form solid. There must be reduced heat and pressure. The deposition of gas into solid occurs through the removal of thermal energy. The air looses thermal energy and changes into solid.Which is a statement of cell theory? All cells are made up of living molecules. All plants are made of cells. All animals are made of cells. All cells are produced from other cells.
Answer:
all cells are produced from other preexisting cells through cell division
When 1 mole of CO(g) reacts with H2O(l) to form CO2(g) and H2(g) according to the following equation, 2.80 kJ of energy are absorbed. CO(g) + H2O(l)CO2(g) + H2(g) Is this reaction endothermic or exothermic? _________ What is the value of q? kJ
Elimination of the pharmaceutical IV antibiotic gentamicin follows first-order kinetics. If the half-life of gentamicin is 1.5 hours for an adolescent. What fraction of the original reactant concentration will remain after 8 hours if the original concentration was 8.4 x 10-5 M.
Explanation:
The given data is:
The half-life of gentamicin is 1.5 hrs.
The reaction follows first-order kinetics.
The initial concentration of the reactants is 8.4 x 10-5 M.
The concentration of reactant after 8 hrs can be calculated as shown below:
The formula of the half-life of the first-order reaction is:
[tex]k=\frac{0.693}{t_1_/_2}[/tex]
Where k = rate constant
t1/2=half-life
So, the rate constant k value is:
[tex]k=\frac{0.693}{1.5 hrs}[/tex]
The expression for the rate constant is :
[tex]k=\frac{2.303}{t} log \frac{initial concentration}{concentration after time "t"}[/tex]
Substitute the given values and the k value in this formula to get the concentration of the reactant after time 8 hrs is shown below:
[tex]\frac{0.693}{1.5 hrs} =\frac{2.303}{8 hrs} x log \frac{8.4x10^-^5}{y} \\ log \frac{8.4x10^-^5}{y} =1.604\\\frac{8.4x10^-^5}{y}=10^1^.^6^0^4\\\frac{8.4x10^-^5}{y}=40.18\\y=\frac{8.4x10^-^5}{40.18} \\=>y=2.09x10^-^6[/tex]
Answer: The concentration of reactant remains after 8 hours is 2.09x10^-6M.
How many grams of glucose are needed to prepare 144.3 mL of a 1.4%(m/v) glucose solution?
Answer:
2.0202 grams
Explanation:
1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.
so ?g glucose = 144.3 mL soln
Now apply the conversion factor, and you have:
?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).
so you have (144.3x1.4/100) g glucose= 2.0202 grams
what type of bonding does Sodium Sulphate comes under?and explain in detail please
Answer:
The bond between sodium sulfate is an ionic bond since it's a bond between a metal and non metals however the bond between sulfur and oxygen is a covalent bond since the two are non metals and the other reason that makes this an ionic bond is that there is both losing and gaining of electrons..
I hope this helps
9. Discuss the general trend in Chemical Properties of the Representative Elements
Answer:
Elements in the same period show trends in atomic radius, ionization energy, electron affinity, and electronegativity.
A metal (C = 0.2158 cal/g· °C) is removed from a hot (350. °F) oven in which it had achieved thermal equilibrium. The metal is placed into 200. mL acetic acid. The temperature of the acid increases to 90.3 °C from 24.3 °C. What is the mass of the metal? (dacetic acid = 1.04 g/cm3; Cs, acetic acid = 2.055 J/g·°C) Group of answer choices 120. g 362 g 1452 g 347 g 281 g
Answer:
362g
Explanation:
heat lost by metal= heat gained by acetic acid
tfs are the same so you cando delta T
convert Cal/gc to J/gc
thectgod ig follow
What is alkaline and what is acidic pH
Answer:
An alkaline is a substance that dissolves in water to produce hydroxyl ions (OH-)
Explanation:
The pH range of an alkaline is from 8–14.
Acidic pH ranges from 0–6.9.
Complete the balanced dissociation equation for the compound below in aqueous solution. If the compound does not dissociate, write NR after the reaction arrow.
HI (aq) -->
Answer:
[tex]{ \bf{HI _{(aq)} \: → \: H {}^{ + } _{(aq)} \: + \: \: I {}^{ - } _{(aq)} }}[/tex]
Which species is the conjugate base of H2SO3
Explanation:
As you know, the conjugate base of an acid is determined by looking at the compound that's left behind after the acid donates one of its acidic hydrogen atoms.
The compound to which the acid donates a proton acts as a base. The conjugate base of the acid will be the compound that reforms the acid by accepting a proton.
In this case, sulfurous acid has two protons to donate. However, the conjugate base of sulfurous acid will be the compound left behind after the first hydrogen ion is donated.
There are _______ alkanes with molecular formula C10H22
a. 74
b. 75
c. 76
d. 77
Cathodic protection of iron involves using another more reactive metal as a sacrificial anode. Classify each of the following metals by whether they would or would not act as a sacrificial anode to iron under standard conditions.
a. Ag
b. Mg
c. Cu
d. Pb
e. Sn
f. Zn
g. Au
Answer:
a. Ag ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
b. Mg ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.
c. Cu ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
d. Pb ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
e. Sn ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
f. Zn ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.
g. Au ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
Explanation:
Cathodic protection of iron involves using another more reactive metal as a sacrificial anode. The reactivity series of metals arranges metals based on decreasing order of reactivity. The more reactive metals are found higher up in the series while the least reactive metals are found at the lower ends of the series. Thus, metals above iron in the reactivity series can serve as sacrificial anodes by protecting against corrosion, while those lower than iron cannot.
Based on the reactivity series, the following metals can be classified as either a sacrificial anode for iron or not:
a. Ag ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
b. Mg ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.
c. Cu ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
d. Pb ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
e. Sn ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
f. Zn ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.
g. Au ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.
In an analysis of interhalogen reactivity, 0.350 mol ICl was placed in a 5.00 L flask and allowed to decompose at a high temperature.
2 ICl(g) I2(g) + Cl2(g)
Calculate the equilibrium concentrations of I2, Cl2, and ICl. (Kc = 0.110 at this temperature.)
I2 M
Cl2 M
ICl M
Answer:
[ICl] = 0.0420 M
[I₂] = [Cl₂] = 0.0140 M
Explanation:
Step 1: Calculate the initial concentration of ICl
[ICl] = 0.350 mol / 5.00 L = 0.0700 M
Step 2: Make an ICE chart
2 ICl(g) ⇄ I₂(g) + Cl₂(g)
I 0.0700 0 0
C -2x +x +x
E 0.0700-2x x x
The concentration equilibrium constant (Kc) is:
Kc = 0.110 = [I₂] [Cl₂] / [ICl]² = x² / (0.0700-2x)² = (x/0.0700-2x)²
0.332 = x/0.0700-2x
x = 0.0140
The concentrations at equilbrium are:
[ICl] = 0.0700-2x = 0.0700-0.0280 = 0.0420 M
[I₂] = [Cl₂] = x = 0.0140 M
What minimum mass of HCl in grams would you need to dissolve a 2.2 g iron bar on a
padlock?
2.8 g is the minimum mass of HCl in grams that would you need to dissolve a 2.2 g iron bar on a padlock.
What is dissolution?When a solute is dissolved in a solvent, a solution is created. Dissolution is the process through which solutes, or dissolved parts, combine to form a solution inside a solvent. In this procedure, the gas, liquid, or solid dissolves inside the original solvent and forms a solution.
In some polymer applications, dissolution is also an issue since it results in swelling, a loss of strength and stiffness, and a change in volume. Whether a chemical process is man-made or natural, dissolution is crucial. Catalysts are tested using dissolution. 2.8 g is the minimum mass of HCl in grams that would you need to dissolve a 2.2 g iron bar on a padlock.
Therefore, 2.8 g is the minimum mass of HCl in grams that would you need to dissolve a 2.2 g iron bar on a padlock.
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Preparation the buffer solution: initial pH of buffer solution: ____ Titration of a weak acid with a strong base: initial pH of weak acid: ____ final pH of weak acid: ____ Amount of NaOH added: ____ Titration Curve for Weak Acid with a Strong Base (Paste curve here.)
Answer:
pH of buffer solution is 7.0
Initial pH of Weak acid is 3.27
Final pH of weak acid is 3.07
Amount of NaOH added is 1ml
Explanation:
Titration is a process in which acid and base are introduced together until a neutral solution is achieved whose pH value is near to buffer solution which is 7.0, the pH value for acid is below 7 while pH value for base is above 7.
Aspirin that has been stored for a long time may give a vinegar like odour and give a purple colour with FeCl3. What reaction would cause this to happen
?.
Answer:
See explanation
Explanation:
The IUPAC name of aspirin is 2-Acetoxybenzoic acid. It is composed of an acetoxy moiety and a benzoic acid moiety.
The compound can be hydrolysed under prolonged storage conditions to yield acetic acid which causes the vinegar like odour.
Also, one of the products of this hydrolysis bears a phenol group which reacts with FeCl3 to give a purple color.
The enthalpy of vaporization of water is 2,257,000 J/kg. If I have a 1 kg sample, how much energy is needed to boil all of it
Answer:
2257000 J
Explanation:
Applying,
Q = Cₓm.................. Equation 1
Where Q = amount of energy need to boil the water, Cₓ = Enthalpy of vaporization of water, m = mass of water.
From the question,
Given: Cₓ = 2257000 J/kg, m = 1 kg
Substitute these values into equation 1
Q = 2257000×1
Q = 2257000 J
Hence the energy needed to boil all of the water is 2257000 J
Too many objects inside a laboratory fume hood can disrupt the airflow and possibly compromise you safety. Which of the following are considered best practices in the use of a laboratory fume hood?
a. Open the sash as much as possible
b. Work at least 25 cm inside the hood
c. Use fast, quick movements to limit your exposure
d. Place objects to one side—work on other side
e. Use a raised along the back of the hood
Best practices for fume hoods: work 25 cm inside, organize items to one side, use raised ledge; avoid open sash and quick movements.
Laboratory fume hoods must be used safely. Workers should operate at least 25 cm within the hood to preserve ventilation and avoid dangerous chemicals. Place things on one side of the hood to preserve ventilation and prevent clogging.
A raised ledge on the rear of the hood prevents things from falling in and impeding airflow. Avoid fully opening the sash to maintain ventilation and containment. Fast, rapid motions can interrupt airflow, so prevent them. These practises guarantee the fume hood contains harmful compounds, making the lab safer. Therefore, option (B), (D) and (E) are correct.
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Once you have collected 40 mL of distillate, you should ________. turn off your hot plate lower your lab jack carelessly use your hand to remove the heating block turn off the hot plate and carefully lower the lab jack, making sure that no cords or hoses get caught in it
Answer:
Once you have collected 40 mL of distillate, you should ________.
turn off the hot plate and carefully lower the lab jack, making sure that no cords or hoses get caught in it.
Explanation:
Distillate is the product obtained from the process of distillation. Distillation is the separation of components of a liquid mixture based on different boiling points. Distillation can be used to purify alcohol, for desalination, refining of crude oil, and for obtaining liquefied gases. A lab jack is an essential tool that supports and lifts hotplates, glassware, baths, and other small lab equipment requiring stable surfaces at a specific height.
FORMULAS OF IONIC COMPOUNDS
FIND: POSITIVE ION, NEGATIVE ION AND FORMULA IN:
NAME:
Sodium chloride
Magnesium chloride
Calcium oxide
Lithium phosphide
Aluminum sulfide
Calcium nitride
Iron(III)chloride
Iron(II)oxide
Copper(I)sulfide
Copper(II)nitride
Zinc oxide
Silver sulfide
Potassium carbonate
Sodium nitrate
Calcium bicarbonate
Aluminum hydroxide
Lithium phosphate
Potassium sulfate
Answer:
NaCl, Na⁺,Cl⁻.
MgCl₂, Mg²⁺, Cl⁻.
CaO, Ca²⁺, O²⁻.
Li₃P, Li⁺, P³⁻.
Al₂S₃, Al³⁺, S²⁻.
Ca₃N₂, Ca²⁺, N³⁻.
FeCl₃, Fe³⁺, Cl⁻.
FeO, Fe²⁺, O²⁻.
Cu₂S, Cu⁺, S²⁻.
Cu₃N₂, Cu²⁺, N³⁻.
ZnO, Zn²⁺, O²⁻.
Ag₂S, Ag⁺, S²⁻.
K₂CO₃, K⁺, CO₃²⁻.
NaNO₃, Na⁺, NO₃⁻.
Ca(HCO₃)₂, Ca²⁺, HCO₃⁻.
Al(OH)₃, Al³⁺,OH⁻.
Li₃PO₄, Li⁺, PO₄³⁻.
K₂SO₄, K⁺, SO₄²⁻.
Explanation:
Sodium chloride. NaCl, formed by the cation Na⁺ and the anion Cl⁻.
Magnesium chloride. MgCl₂, formed by the cation Mg²⁺ and the anion Cl⁻.
Calcium oxide. CaO, formed by the cation Ca²⁺ and the anion O²⁻.
Lithium phosphide. Li₃P, formed by the cation Li⁺ and the anion P³⁻.
Aluminum sulfide. Al₂S₃, formed by the cation Al³⁺ and the anion S²⁻.
Calcium nitride. Ca₃N₂, formed by the cation Ca²⁺ and the anion N³⁻.
Iron(III)chloride. FeCl₃, formed by the cation Fe³⁺ and the anion Cl⁻.
Iron(II)oxide. FeO, formed by the cation Fe²⁺ and the anion O²⁻.
Copper(I)sulfide. Cu₂S, formed by the cation Cu⁺ and the anion S²⁻.
Copper(II)nitride. Cu₃N₂, formed by the cation Cu²⁺ and the anion N³⁻.
Zinc oxide. ZnO, formed by the cation Zn²⁺ and the anion O²⁻.
Silver sulfide. Ag₂S, formed by the cation Ag⁺ and the anion S²⁻.
Potassium carbonate. K₂CO₃, formed by the cation K⁺ and the anion CO₃²⁻.
Sodium nitrate. NaNO₃, formed by the cation Na⁺ and the anion NO₃⁻.
Calcium bicarbonate. Ca(HCO₃)₂, formed by the cation Ca²⁺ and the anion HCO₃⁻.
Aluminum hydroxide. Al(OH)₃, formed by the cation Al³⁺ and the anion OH⁻.
Lithium phosphate. Li₃PO₄, formed by the cation Li⁺ and the anion PO₄³⁻.
Potassium sulfate. K₂SO₄, formed by the cation K⁺ and the anion SO₄²⁻.
explain why it is important not to correct any gas from the first few seconds of the experiment
Answer:
gu kha fuschhehdjdvdbeodbr
What mass of NaNO3 must be dissolved to make 838mL of a 1.25 M solution
Answer:
89.04 g of NaNO₃.
Explanation:
We'll begin by converting 838 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
838 mL = 838 mL × 1 L / 1000 mL
838 mL = 0.838 L
Next, we shall determine the number of mole of NaNO₃ in the solution. This can be obtained as follow:
Volume = 0.838 L
Molarity = 1.25 M
Mole of NaNO₃ =?
Mole = Molarity × volume
Mole of NaNO₃ = 1.25 × 0.838
Mole of NaNO₃ = 1.0475 mole
Finally, we shall determine the mass of NaNO₃ needed to prepare the solution. This can be obtained as follow:
Mole of NaNO₃ = 1.0475 mole
Molar mass of NaNO₃ = 23 + 14 + (16×3)
= 23 + 14 + 48
= 85 g/mol
Mass of NaNO₃ =?
Mass = mole × molar mass
Mass of NaNO₃ = 1.0475 × 85
Mass of NaNO₃ = 89.04 g
Therefore, 89.04 g of NaNO₃ is needed to prepare the solution.
The functional groups in an organic compound can frequently be deduced from its infrared absorption spectrum.
a. True
b. False
Answer:
a. True
Explanation:
The main information that gives an infrared absorption spectrum is the type of functional groups that are present in an organic compound. The infrared (IR) spectroscopy is based on the fact that functional groups absorb light in the IR region of the electromagnetic spectrum (approximately at 2,500-16,000 nm) and induces a vibrational excitation of the covalently bonded atoms in the group. The vibration of the atoms can be of different types, such as stretching, bending, etc. Each functional group (such as the carbonyl group) in an organic compound absorbs at a specific IR frequency so they can be distinguished from an IR spectrum.
Convert 1.25 x 1024 atoms of carbon to moles of carbon.
Answer:
2.076
Explanation:
1 mole is 6.02 * 10^23
To convert from atoms (or molecules or compounds or ions etc.) to mols, you divide the number of atoms (or molecules or etc.) by 6.02 * 10^23
So it is (1.25 * 10^24)/(6.02 * 10^23)
=2.076
Answer:
[tex]\boxed {\boxed {\sf 2.08 \ mol \ C}}[/tex]
Explanation:
We are asked to convert a number of carbon atoms to moles.
We will use Avogadro's Number for this, which is 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. For this problem, the particles are atoms of carbon. There are 6.022 ×10²³ atoms of carbon in 1 mole of carbon.
We will also use dimensional analysis to solve this problem. To do this, we use ratios. Set up a ratio using the underlined information.
[tex]\frac {6.022 \times 10^{23} \ atoms \ C}{1 \ mol \ C}[/tex]
We are converting 1.25 ×10²⁴ atoms of carbon to moles, so we multiply the ratio by that value.
[tex]1.25 \times 10^{24} \ atoms \ C* \frac {6.022 \times 10^{23} \ atoms \ C}{1 \ mol \ C}[/tex]
Flip the ratio. It remains equivalent, but it allows us to cancel the units atoms of carbon.
[tex]1.25 \times 10^{24} \ atoms \ C* \frac{1 \ mol \ C} {6.022 \times 10^{23} \ atoms \ C}[/tex]
[tex]1.25 \times 10^{24} * \frac{1 \ mol \ C} {6.022 \times 10^{23} }[/tex]
[tex]\frac{1.25 \times 10^{24} } {6.022 \times 10^{23} } \ mol \ C[/tex]
[tex]2.075722351 \ mol \ C[/tex]
The original measurement of atoms has three significant figures, so our answer must have the same. For the number we calculated, that is the hundredths place. The 5 in the thousandths place tells us to round the 7 up to an 8.
[tex]2.08 \ mol \ C[/tex]
1.25 ×10²⁴ atoms of carbon is equal to approximately 2.08 moles of carbon.
