CHEMISTRY COLLEGE

What volume of 1.27 M HCl is required to prepare 197.4 mL of 0.456 M HCl

Answers

Answer 1

Answer:

Answer:

70.88 mL volume of 1.27 M of HCl is required.

Explanation:

Given data:

Initial volume = ?

Initial molarity = 1.27 M

Final volume = 197.4 mL

Final molarity = 0.456 M

Solution:

Formula:

M₁V₁ = M₂V₂

Now we will put the values in formula.

1.27 M × V₁ = 0.456 M × 197.4 mL

V₁ = 0.456 M × 197.4 mL/1.27 M

V₁ = 90.014M.mL/1.27 M

V₁ = 70.88 mL

70.88 mL volume of 1.27 M of HCl is required.

Answer 2

Answer:

Final answer:

To prepare 197.4 mL of 0.456 M HCl from 1.27 M HCl, you need 71.03 mL of 1.27 M HCl.

Explanation:

The subject of this problem involves using the concept of molarity in Chemistry. We can use a simple formula for dilution, M1V1 = M2V2, to find the volume. Here M1 (1.27 M) is the molarity of stock HCl, V1 is the required volume, M2 (0.456 M) is the desired molarity, and V2 (197.4 mL) is the volume of the solution. Solving for V1, we get V1 = M2V2 / M1 = (0.456 M * 197.4 mL) / 1.27 M = 71.03 mL. Therefore, 71.03 mL of 1.27 M HCl is needed to prepare 197.4 mL of 0.456 M HCl.

Learn more about Molarity Calculations here:

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1. On the basis of your results, what is the relationship between the temperature of the solvent and the rate of solution formation?

Answers

Answer: As the temperature increases, the rate of formation of solution increases.

Explanation:

Solvent is defined as a substance which is present in larger proportion in a solution. Solute is dissolved in the solvent to form a solution.

As, the temperature of the solvent increases, the kinetic energy of the particles of solvent increases and the intermolecular spacing between the solvent particles increases and therefore, this results in the more dissolution of the solute particles in the solvent and hence, the formation of solution increases.

Therefore, there is a direct relationship between the temperature of the solvent and the formation of the solution.

Answer:

Hot temp = solution forming faster.

Explanation:

Salt in hot water dissolves faster than salt in ice water.

Water contains 2 polar bonds and the molecule is polar
True
False

Answers

TRUEEEEEEEEEEEEEEEEEEEEEEE

Place whole number coefficients in the blanks to balance the chemical reaction. ____ KOH + ____ AlCl3 --> _____ Al(OH)3 + _____KCl

Answers

3KOH +ALCL3》 Al(OH)3+ 3KCL

The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentration of 0.086 M, calculate the concentration of NOBr after 22 s. (b) Calculate the half-lives when [NOBr]0 5 0.072 M and [NOBr]0 5 0.054 M.

Answers

Explanation:

2NOBr(g) --> 2NO(g) 1 Br2(g)

Rate constant, k = 0.80

a) Initial concentration, Ao = 0.086 M

Final Concentration, A = ?

time = 22s

These parameters are connected with the equation given below;

1 / [A] = kt + 1 / [A]o

1 / [A] = 1 / 0.086 + (0.8 * 22)

1 / [A] = 11.628 + 17.6

1 / [A] = 29.228

[A] = 0.0342M

b) t1/2 = 1 / ([A]o * k)

when [NOBr]0 5 0.072 M

t1/2 = 1 / (0.072 * 0.80)

t1/2 = 1 / 0.0576 = 17.36 s

when [NOBr]0 5 0.054 M

t1/2 = 1 / (0.054 * 0.80)

t1/2 = 1 / 0.0432 = 23.15 s

Answer:

(a)

$0.0342M$

(b)

$t_(1/2)=17.36s\nt_(1/2)=23.15s$

Explanation:

Hello,

(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:

$(1)/([NOBr])=kt +(1)/([NOBr]_0)\n(1)/([NOBr])=(0.8)/(M*s)*22s+(1)/(0.086M)=(29.3)/(M)\n$

$[NOBr]=(1)/(29.2/M)=0.0342M$

(b) Now, for a second-order reaction, the half-life is computed as shown below:

$t_(1/2)=(1)/(k[NOBr]_0)$

Therefore, for the given initial concentrations one obtains:

$t_(1/2)=(1)/((0.80)/(M*s)*0.072M)=17.36s\nt_(1/2)=(1)/((0.80)/(M*s)*0.054M)=23.15s$

Best regards.

Systematic name for: Silver Nitrate.molar mass of Ag:
molar mass of N:
molar mass of O:

and the overall molar mass for Silver Nitrate.

Answers

Answer:

$AgNO_3$ ,Molar mass =169.87 g/mol

Explanation:

The systematic name for silver nitrate is $AgNO_3$

Now we have to calculate the molar mass of $AgNO_3$

Molar mass of Ag = 107.87 g/mol

Molar mass of N =14 g/mol

Molar mass of O =16 g/mol

So the molar mass of silver nitrate ( $AgNO_3$ ) is

$=(1* 107.87+1* 14+3* 16)=169.87g/mol$

What is the mole fraction of calcium chloride in 3.35 m CaCl2(aq)? The molar mass of CaCl2 is 111.0 g/mol and the molar mass of water is 18.02 g/mol.

Answers

Answer: The mole fraction of calcium chloride and water in the solution is 0.057 and 0.943 respectively

Explanation:

We are given:

Molality of calcium chloride = 3.35 m

This means that 3.35 moles of calcium chloride are present in 1 kg or 1000 g of water

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of water = 1000 g

Molar mass of water = 18.02 g/mol

Putting values in above equation, we get:

$\text{Moles of water}=(1000g)/(18.02g/mol)=55.49mol$

Total moles of solution = [3.35 + 55.49] = 58.84 moles

Mole fraction of a substance is given by:

$\chi_A=(n_A)/(n_A+n_B)$

For calcium chloride:

$\chi_(CaCl_2)=(n_(CaCl_2))/(n_(CaCl_2)+n_(H_2O))\n\n\chi_(CaCl_2)=(3.35)/(58.84)=0.057$

For water:

$\chi_(H_2O)=(n_(H_2O))/(n_(CaCl_2)+n_(H_2O))\n\n\chi_(H_2O)=(55.49)/(58.84)=0.943$

Hence, the mole fraction of calcium chloride and water in the solution is 0.057 and 0.943 respectively

Answer:

49.3% water

Explanation:

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