how do you Calculate the molarity of 250 ml of solution in which 2.7 g of MgCl2 are dissolved

Answer:

Water filters remove elements that cause drinking water to have an unpleasant taste and smell, such as lead, chlorine and bacteria. Home water filtration system will improve the overall purity, taste and smell of your drinking water. It also lowers the pH level of the water that you drink.

Explanation:

here's the answer to your question

Answer:

LiOH(aq) → Li⁺(aq) + OH⁻(aq). 

Answer:

26mL

Explanation:

NaOH+HCl= NaCl+H2O

nHCl=0.0243*2=0.0486

nNaOh=nHCl

VNaOH=0.0486/1.87=0.026l=26ml

Answer:

26.0 mL

Explanation:

Answer:

a. 4.41 g of Urea

b. 1.5 g of Urea

Explanation:

To start the problem, we define the reaction:

2NH₃ (g) +  CO₂ (g) → CH₄N₂O (s)  +  H₂O(l)

We only have mass of ammonia, so we assume the carbon dioxide is in excess and ammonia is the limiting reactant:

2.6 g . 1mol / 17g = 0.153 moles of ammonia

Ratio is 2:1. 2 moles of ammonia can produce 1 mol of urea

0.153 moles ammonia may produce, the half of moles

0153 /2 = 0.076 moles of urea

To state the theoretical yield we convert moles to mass:

0.076 mol . 58 g/mol = 4.41 g

That's the 100 % yield reaction

If the percent yield, was 34%:

4.41 g . 0.34 = 1.50 g of urea were produced.

Formula is (Yield produced / Theoretical yield) . 100 → Percent yield

Answer:

I say Star S has radial motion

Explanation:

I'm not sure if it right but let me know if you have any other questions

Answer:

21.5mL of a 0.100M HCl are required

Explanation:

The sodium phenoxide reacts with HCl to produce phenol and NaCl in a 1:1 reaction.

To solve this question we need to find the moles of sodium phenoxide. These moles = Moles of HCl required to reach equivalence point and, with the concentration, we can find the needed volume as follows:

Mass NaC6H5O:

1.000g * 25% = 0.250g NaC6H5O

Moles NaC6H5O -116.09g/mol-

0.250g NaC6H5O * (1mol/116.09g) = 2.154x10⁻³ moles = Moles of HCl required

Volume 0.100M HCl:

2.154x10⁻³ moles HCl * (1L/0.100mol) = 0.0215L =

Answer:

Li^3+

Explanation:

The electronic configuration of lithium is ; 1s2 2s1. This means that lithium has one electron in its outermost shell and two core electrons.

We know that it is difficult to remove these core electrons during ionization. Lithium belongs to group 1 hence Li^+ is formed more easily.

It is very difficult to form Li^3+ because it involves loss of core electrons which requires a lot of energy.

Answer:

However, there has to be 2 electrons on the first shell, and 8 on the others.

Explanation:

Hope this helps :)

Answer:

19.2 g

Explanation:

Step 1: Write the balanced equation

P₂O₅(s) + 3 H₂O(l) ⇒ 2 H₃PO₄

Step 2: Calculate the moles corresponding to 5.29 g of H₂O

The molar mass of H₂O is 18.02 g/mol.

5.29 g × 1 mol/18.02 g = 0.294 mol

Step 3: Calculate the theoretical yield of phosphoric acid, in moles

The molar ratio of H₂O to H₃PO₄ is 3:2. The theoretical yield of H₃PO₄ is 2/3 × 0.294 mol = 0.196 mol

Step 4: Calculate the mass corresponding to 0.196 moles of H₃PO₄

The molar mass of H₃PO₄ is 97.99 g/mol.

0.196 mol × 97.99 g/mol = 19.2 g

After the double replacement reaction is complete 0 grams of BaCl₂ and 31.16 grams of H₂SO₄ will remain.

First, we will write the balanced equation for the reaction

H₂SO₄ + BaCl₂  → BaSO₄ + 2HCl

This means 1 mole of BaCl₂ is needed to react completely with 1 mole of H₂SO₄ to give 1 mole of BaSO₄ and 2 moles of HCl

From the question, 50.0g of sulfuric acid is mixed with 40.0 grams of barium chloride. To determine the quantity of each substance remaining after the complete reaction, we will first determine the number of moles present in each of the reactant.

For H₂SO₄

mass = 50.0g

Molar mass = 98.079 g/mol

From the formula

Number of moles = Mass / Molar mass

∴ Number of moles of H₂SO₄ = 50.0g / 98.079 g/mol

Number of moles of H₂SO₄ = 0.5098 mol

For BaCl₂

mass = 40.0 g

Molar mass = 208.23 g/mol

∴ Number of moles of BaCl₂ = 40.0g / 208.23 g/mol

Number of moles of BaCl₂ = 0.1921 mol

Since the number of moles of H₂SO₄ is more than that of BaCl₂, then H₂SO₄ is the excess reagent and BaCl₂ is the limiting reagent (that is, it will be used up completely during the reaction)

From the equation, 1 mole of H₂SO₄ is needed to completely react with 1 mole of BaCl₂

∴ 0.1921 mol of H₂SO₄ will be needed to completely react with 0.1921 mol of BaCl₂.

Therefore, after the reaction is complete, 0 mole (i.e 0 grams) of BaCl₂ will remain and (0.5098 mole - 0.1921 mole) of H₂SO₄ will remain.

Number of moles H₂SO₄ that will remain = 0.5098 mole - 0.1921 mole = 0.3177 moles

Now, we will convert this to grams

From the formula

Mass = Number of moles × Molar mass

Mass of H₂SO₄ that will remain = 0.3177 moles × 98.079 g/mol

Mass of H₂SO₄ that will remain = 31.1597 g

Mass of H₂SO₄ that will remain ≅ 31.16 g

Hence, after the double replacement reaction is complete 0 grams of BaCl₂ and 31.16 grams of H₂SO₄ will remain.

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Answer:

I think its A maybe am not sure

Answer:

See detailed explanation.

Explanation:

Hey there!

In this case, according to the given information, it turns out possible for us to solve this problem by firstly calculating the moles of each element, assuming those percentages are masses, so that we divide by their molar masses:

[tex]C=\frac{18.63}{12.01}=1.55\\\\H=\frac{1.56}{1.01} =1.55\\\\O=\frac{24.82}{16.00}=1.55\\\\Cl=\frac{54.99}{35.45}=1.55[/tex]

Then, we divide all of them by 1.55 to realize the empirical formula is:

[tex]CHOCl[/tex]

Whose molar mass is 64.47 g/mol, and therefore, since the molar mass of these two is the same, we infer the molecular formula is also CHOCl.

The Lewis structure is shown on the attached document, whereas, the central atom is C and it does complete its octet as well as both O and Cl.

Regards!

Answer:

A) XY3

In both of the atom, they want to have 8 valence electrons on the outer shell so they can become stable.

Answer:

1.28 M

Explanation:

Step 1: Given data

Mass of luminol (solute): 17.0 g

Volume of water: 75.0 mL (this is also the volume of solution)

Step 2: Calculate the moles corresponding to 17.0 g of luminol

The molar mass of luminol is 177.16 g/mol.

17.0 g × 1 mol/177.16 g = 0.0960 mol

Step 3: Calculate the molarity of the solution

We will use the definition of molarity

M = moles of solute / liters of solution

M = 0.0960 mol / 0.0750 L = 1.28 M

Ionization potential and ionization energy are two terms used to describe the same thing.

The ionization potential of hydrogen atom is 13.6 eV

The ionization potential is the energy that is required to remove an electron from the neutral atom. It is the same as the ionization energy.

From the question, we can see that the ionization energy of the hydrogen atom is 13.6 eV, it also means that the ionization potential of the hydrogen atom is also 13.6 eV.

Therefore, If ionization energy of hydrogen atom is 13.6 eV then its ionization potential will be 13.6 eV

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Answer:

hello? are you still here? reply if you are

Answer:

[tex]M_{CO_2}= 25.7g[/tex]

Explanation:

From the question we are told that:

Temperature [tex]T=27.0[/tex]

Volume [tex]V=30L[/tex]

Pressure [tex]P=0.480atm[/tex]

Generally the equation for Ideal gas is mathematically given by

PV=nRT

Therefore

[tex]n=\frac{0.480 x 30}{0.08205 x 300}[/tex]

[tex]n=0.59moles[/tex]

Generally Mass of CO2 is given as

[tex]M_{CO_2}= 0.59 * 44 g/mol[/tex]

[tex]M_{CO_2}= 25.7g[/tex]

Answer:

a Leaning towards the speaker

Answer:

is this question or you just asking I can't understand.

Answer:

[tex]T_2= 36.7 \textdegree C[/tex]

Explanation:

Mass of Water [tex]m_w=6.90kg[/tex]

Temperature [tex]T=34.7 degrees[/tex]

Heat Flow [tex]H=57.1kJ[/tex]

Specific heat capacity of water [tex]\mu= 4.18J.g^(-1).K^(-1)[/tex]

Generally the equation for Final Temperature is mathematically given by

[tex]M*\mu *T_1 + Q = M*\mu *T_2[/tex]

[tex]T_2=\frac{M*\mu *T_1 + Q }{M*\mu}[/tex]

Therefore

[tex]T_2=\frac{6.90*4.18*34.7 + 57.1}{6.90*4.18}[/tex]

[tex]T_2= 36.7 \textdegree C[/tex]

Answer:

a. -58 millivolts

Explanation:

The given Nernst equation is:

[tex]E_{ion} = 58 millivolts /z \Big[ log_{10} \Big( \dfrac{[ion]_{out}}{[ion]_{in}}\Big) \Big]}[/tex]

The equilibrium potential given by the Nernst equation can be determined by using the formula:

[tex]E_{Cl^-} = \dfrac{2.303*R*T}{ZF} \times log \dfrac{[Cl^-]_{out}} {[Cl^-]_{in}}[/tex]

where:

gas constant(R) = 8.314 J/K/mol

Temperature (T) = (20+273)K

= 298K

Faraday constant F = 96485 C/mol

Number of electron on Cl = -1

[tex]E_{Cl^-} = \dfrac{2.303*8.314*298} {(-1)*(96845)} \times log \dfrac{100} {10}[/tex]

[tex]E_{Cl^-} = - 0.05814 \ volts[/tex]

[tex]\mathsf{E_{Cl^-} = - 0.05814 \times 1000 \ milli volts}[/tex]

[tex]\mathsf{E_{Cl^-} \simeq - 58\ milli volts}[/tex]

Answer:

a. 1 mole of acid is equal to one equivalent.

b. 1.00 moles of HCl are found.

c. 1L of 2.00M NaOH is needed to reach the equivalence point

Explanation:

HCl reacts with NaOH as follows:

HCl + NaOH → NaCl + H2O

Where 1 mole of HCl reacts with 1 mole of NaOH. The reaction is 1:1

a. As the reaction is 1:1, 1 mole of acid is equal to one equivalent

b. The initial moles of HCl are:

1.00L * (2.00moles HCl / 1L) = 2.00 moles of HCl

At the halfway point, the moles of HCl are the half, that is:

1.00 moles of HCl are found

c. At equivalence point, we need to add the moles of NaOH needed for a complete reaction of the moles of HCl. As the moles of HCl are 2.00 and the reaction is 1:1, we need to add 2.00 moles of NaOH, that is:

2.00moles NaOH * (1L / 2.00mol) =

1L of 2.00M NaOH is needed to reach the equivalence point

Explanation:

the force of the lone pairs from the bottom would cancel out the force of the lone pairs from the top. Thus, the molecule will be linear.

Answer:

C

Explanation:

Answer:

[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.

Explanation:

Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.

Analysis:

            H₂CO₃(aq)     ⇄     H⁺(aq)    +    HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷

C(i)          0.115M                      0                  0

ΔC              -x                        +x                  +x

C(eq)    0.115M - x                   x                    x

            ≅ 0.115M

Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M

= 4.3 x 10⁻⁷  => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.

In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion  concentration, the hydroxide ion concentration is then calculated from

[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.

________________________________________________________

NOTE: The 2.32 x 10⁻⁴M  value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.

The concentration of hydroxide ion of given solution is 4.3 x 10⁻¹¹M.

We can calculate the concentration of hydroxide ions as follow:

[OH⁻][H⁺] = 10⁻¹⁴

Given chemical reaction with ICE table shown as below:

                H₂CO₃(aq)     ⇄     H⁺(aq)    +    HCO₃⁻(aq)

Initial:             0.115                      0                    0

Change:           -x                        +x                  +x

Equilibrium:  0.115-x                   +x                  +x

Given that, Ka = 4.3 x 10⁻⁷

Equilibrium constant for this reaction is written as:

Ka = [H⁺][HCO₃⁻]/[H₂CO₃]

4.3 x 10⁻⁷ = x² / 0.115

x = 2.32 x 10⁻⁴M = [H⁺]

Now we calculate the concentration of hydroxide ion as:

[OH⁻][H⁺] = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ / 2.32 x 10⁻⁴ = 4.3 x 10⁻¹¹M

Hence, value of [OH⁻] is 4.3 x 10⁻¹¹M.

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a].When sand is added to water it either hangs in the water or forms a layer at the bottom of the container. Sand therefore does not dissolve in water and is insoluble. It is easy to separate sand and water by filtering the mixture.

b]. The water filtered by kamal is not safe to drink .

If you run out of water, or cannot carry enough water with you for your entire trip, you may need to source drinking water from natural water sources.

Answer:

[tex]C_t=0.165M[/tex]

Explanation:

From the question we are told that:

Slope [tex]K=0.056 M-1 s -1[/tex]

initial Concentration [tex]C_1=2.2M[/tex]

Time [tex]t=100[/tex]

Generally the equation for Raw law is mathematically given by

[tex]\frac{1}{C}_t=kt+\frac{1}{C}_0[/tex]

[tex]\frac{1}{C}_t=0.056*100+\frac{1}{2.2}_0[/tex]

[tex]C_t=0.165M[/tex]

The second-order reaction is the reaction that depends on the reactants of the first or the second-order reaction. The concentration after 100 seconds will be 0.165 M.

The specific rate constant (k) of the second-order reaction is given in L/mol/s or per M per s. It is the proportionality constant that gives the relation between the concentration and the rate of the reaction.

Given,

Slope (k)= 0.056 per M per s

Initial concentration of the reactant [tex](\rm C_{1})[/tex] = 2.2 M

Time (t) = 100 seconds

The concentration of the reaction after 100 seconds can be given by,

[tex]\rm \dfrac{1}{C_{t}} = kt + \dfrac{1}{C_{1}}[/tex]

Substitute values in the above equation:

[tex]\begin{aligned} \rm \dfrac{1}{C_{t}} &= 0.056 \times 100 + \dfrac{1}{2.02}\\\\&= 0.165 \;\rm M\end{aligned}[/tex]

Therefore, after 100 seconds the concentration is 0.165 M.

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Answer:

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Explanation: