All categories

2 years ago

How many moles of hcl are present in 40.0 ml of a 0.035 m solution?

Chemistry

1 answer:

Elden [556K]2 years ago

8 0

Answer:

0.0014 moles is present in 40cm³ of 0.035M of HCl solution

Explanation:

Molarity = 0.035M

V = 40.0mL

1mL = 1cm³

V = 40cm³

0.035 moles = 1000cm³

X moles is present in 40cm³

X = (40 * 0.035) / 1000

X = 0.0014moles

0.0014 moles is present in 40cm³ of solution

You might be interested in

What happens if your caculating the speed of a longitudinal wave and you increase the temp in the room

nordsb [41]

The speed of a longitudinal wave would increase, as the kinetic energy molecules would move quicker in response to the higher temperature.

5 0

2 years ago

A piece of solid carbon dioxide, with a mass of 7.8g,is placed in a 4.0-L otherwise empty container at 278C. What is the pressur

sukhopar [10]

Answer:

First question: P = 2.00 atm (1520 torr)

Second question: Partial pressure of carbon dioxide: 2.00 atm (1520 torr); total pressure: 2.97 atm (2260 torr).

Explanation:

The molar mass of caron dioxide, CO₂, is:

12 g/mol of C + 2*16 g/mol of O = 44 g/mol

The number of moles is the mass divided by the molar mass:

n = 7.8/44 = 0.1773 mol

By the ideal gas law:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.082 atm*L/mol*K), and T is the temperature (278ºC + 273 = 551 K).

So, the pressure will be:

P*4 = 0.1773*0.082*551

P = 2.00 atm = 1520 torr

If there was air in the container, the partial pressure of the carbon dioxide will still the same, 2 atm (1520 torr), because the partial pressure is the pressure the substance would have if it eas alone in the container.

The total pressure is the sum of the partial pressures, so

Pt = 740 + 1520 = 2260 torr = 2.97 atm

7 0

2 years ago

A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

Ede4ka [16]

<u>Answer:</u> The mass of glucose in final solution is 1.085 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ ......(1)

Given mass of glucose = 15.5 g

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Molarity of glucose solution}=\frac{15.5\times 1000}{180.2\times 100}\\\\\text{Molarity of glucose solution}=0.860M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.860M\\V_1=35.0mL\\M_2=?M\\V_2=0.500L=500mL$

Putting values in above equation, we get:

$0.860\times 35.0=M_2\times 500\\\\M_2=\frac{0.860\times 35.0}{500}=0.0602M$

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of glucose solution = 0.0602 M

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0602=\frac{\text{Mass of glucose solution}\times 1000}{180.2\times 100}\\\\\text{Mass of glucose solution}=\frac{0.0602\times 180.2\times 100}{1000}=1.085g$

Hence, the mass of glucose in final solution is 1.085 grams

4 0

3 years ago

How many moles of ions would you expect in an aqueous solution containing one mole of chromium(III) chloride? Hint: write out th

Cerrena [4.2K]

Answer:

Four

Explanation:

AlCl₃(aq) ⟶ Al³⁺(aq) + 3Cl⁻(aq)

One mole of AlCl₃ produces 1 mol of Al³⁺ and 3 mol of Cl⁻.

That's four moles of ions.

8 0

2 years ago

What do the elements in each pair have in common? k, kr be, mg ni, tc b, ge al, pb?

tensa zangetsu [6.8K]

The pairs are:
K, Kr - Same period
Be, Mg - Same group
Ni, Tc - Both are transition metals
B, Ge - Both are metaloids
Al, Pb - Both form inert oxides

4 0

3 years ago