The graph below shows how the temperature and volume of a gas vary when
the number of moles and the pressure of the gas are held constant. What
happens to the
temperature of a gas as its volume increases?
A. The temperature decreases
B. The temperature increases
C. The temperature remains the same.
D. The temperature doubles.

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.

Answer:

cool or increase pressure

Explanation:

Answer:

a. 6.15 mL b. 30.73 mL

Explanation:

a. What volume of this ketamine solution would the 65.0 kg user have to inject to experience a high at 0.400 mg/kg?

Since we have 0.250 g of ketamine in 1/4 cup of water and 1 cup of water equals 236.5 mL, we need to find the concentration of ketamine we have.

So concentration of ketamine C = mass of ketamine, m/volume of water, V

m = 0.250 g and V = 1/4 cup = 1/4 × 236.5 mL = 59.125 mL

So, C = m/V = 0.250 g/59.125 mL = 0.00423 g/mL = 4.23 mg/mL

Since the user has a mass of 65 kg and requires a high at 0.400 mg/kg, the mass of ketamine for this high is M = 65 kg × 0.400 mg/kg = 26 mg

Since mass, M = concentration ,C × volume, V

M = CV

V = M/C

The volume of ketamine required for the 0.400 mg/kg high is

V = 26 mg/4.23 mg/mL

V = 6.15 mL

b. What volume of this ketamine solution would the user have to inject to become unconscious at 2.00 mg/kg?

Since the concentration of ketamine is C = 4.23 mg/mL, and Since the user has a mass of 65 kg and requires an injection of 2.00 mg/kg to be unconscious, the mass of ketamine required to be unconscious is M' = 65 kg × 2.00 mg/kg = 130 mg

Since mass, M' = concentration ,C × volume, V

M' = CV

V = M/C

The volume of ketamine required for the 2.00 mg/kg unconscious injection is

V = 130 mg/4.23 mg/mL

V = 30.73 mL

The compound in this reaction which is having the elemental ratio of 1:4:1 is NH₄Cl where nitrogen and chlorine are of one mole each with 4 hydrogens.

Elemental ratio of a compound is the ratio of number of atoms  of each  elements in that compound. The elemental ratio can be determined from the molecular formula of compounds.

The given reaction is a double displacement reaction. Here, the Cl group is replaced to the ammonia and OH group is replaced to the water. Thus, two species is replaced in the reaction.

In NH₄Cl, there are one nitrogen, 4 hydrogens and one chlorine atom. Therefore, the elemental ratio of the compound is 1:4:1. The elemental ratio of water is 2:1 and HCl is 1:1 and that in NH₄OH is 1:5:1. Hence, option b is correct.

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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

An isotope is an element with the same atomic number but different mass number due to differences in number of neutrons.

electron configuration is 2,8,6.

Belongs to group 6 and period group 3.

It forms an ion by accepting 2 electrons

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.

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.

Answer:

a. 2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

b. i. NO₂⁻ is the oxidizing agent

ii. NO₃⁻ is the reducing agent.

Explanation:

a. Balance the following skeleton reaction

The reaction is

NO₂ (g) → NO₃⁻ (aq) + NO₂⁻ (aq)

The half reactions are

NO₂ (g) → NO₃⁻ (aq)  (1) and

NO₂ (g) → NO₂⁻  (aq) (2)

We balance the number of oxygen atoms in equation(1) by adding one H₂O molecule to the left side.

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq)

We now add two hydrogen ions 2H⁺ on the right hand side to balance the number of hydrogen atoms

NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq)

The charge on the left hand side is zero while the total charge on the right hand side is -1 + 2 = +1. To balance the charge on both sides, we add one electron to the right hand side.

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻  (4)

Since the number of atoms in equation two are balanced, we balance the charge since the charge on the left hand side is zero and that on the right hand side is -1. So, we add one electron to the left hand side.

So, NO₂ (g) + e⁻ → NO₂⁻  (aq) (5)

We now add equation (4) and (5)

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻  (4)

+  NO₂ (g) + e⁻ → NO₂⁻  (aq) (5)

2NO₂ (g) + H₂O (l) + e⁻ → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq) + e⁻  (4)

2NO₂ (g) + H₂O (l)  → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq)  

We now add two hydroxide ions to both sides of the equation.

So, 2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq) + 2OH⁻ (aq)

The hydrogen ion and the hydroxide ion become a water molecule

2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H₂O (l)

2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

So, the required reaction is

2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

b. Identify the oxidizing agent and reducing agent

Since the oxidation number of oxygen in NO₂ is -2. Since the oxidation number of NO₂ is zero, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = 0

x + 2(-2) = 0

x - 4 = 0

x = 4

Since the oxidation number of oxygen in NO₂⁻ is -1. Since the oxidation number of NO₂⁻ is -1, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = 0

x + 2(-2) = -1

x - 4 = -1

x = 4 - 1

x = 3

Also, the oxidation number of oxygen in NO₃⁻ is -1. Since the oxidation number of NO₃⁻ is -1, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = -1

x + 3(-2) = -1

x - 6 = -1

x = 6 - 1

x = 5

i. The oxidizing agent

The oxidation number of N changes from +4 in NO₂ to +3 in NO₂⁻. So, Nitrogen is reduced and thus  NO₂⁻ is the oxidizing agent

ii. The reducing agent

The oxidation number of N changes from +4 in NO₂ to +5 in NO₃⁻. So, Nitrogen is oxidized and thus and  NO₃⁻ is the reducing agent.

Answer:

all cells are produced from other preexisting cells through cell division

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.

Answer:

11.9 pH

Explanation:

First, we need to find pOH

To find that, we use the formula -log[OH]

-log[7.7x10^-3] = 2.11351

To find the pH, we'll use this formula: 14 = pH + pOH

14 = pH + 2.11351

Subtract boths sides by 2.11351

14 = pH + 2.11351

-2.11351  -2.11351

pH = 11.88649

Answer:

gu kha fuschhehdjdvdbeodbr

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

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.

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

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.

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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Answer:

Elements in the same period show trends in atomic radius, ionization energy, electron affinity, and electronegativity.

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.

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.

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.

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.

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

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

Answer:

2 m/s

Explanation:

Applying the formulae of velocity,

V = d/t............. Equation 1

Where V = Velocity of the body, d = distance, t = time

From the question,

Given: d = 600 m, t = 5 minutes = (5×60) = 300 seconds.

Substitute these values into equation 1

V = 600/300

V = 2 m/s.

Hence the velocity of the body when it travels is 2 m/s

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.

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]

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⁻.