All categories

2 years ago

How many grams of water are needed to dissolve 27.8 g of ammonium nitrate NH4NO3 in order to prepare a 0.452 m solution?

1 answer:

5 0

Answer: 770 g water are needed to dissolve 27.8 g of ammonium nitrate $NH_4NO_3$ in order to prepare a 0.452 m solution

Explanation:

Molality : It is defined as the number of moles of solute present per kg of solvent

Formula used :

$Molality=\frac{n\times 1000}{W_s}$

where,

n= moles of solute

Moles of $NH_4NO_3=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{27.8g}{80.0g/mol}=0.348moles$

$W_s$ = weight of the solvent in g = ?

$0.452=\frac{0.348\times 1000}{W_s}$

$W_s=770g$

Thus 770 g water are needed to dissolve 27.8 g of ammonium nitrate $NH_4NO_3$ in order to prepare a 0.452 m solution

You might be interested in

Coal gasification is a multistep process to convert coal into cleaner-burning fuels. In one step, a coal sample reacts with supe

ddd [48]

Answer :

The enthalpy of reaction is, -187.6 kJ/mol

The total heat will be, -2251 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

(a) The formation of $CH_4$ will be,

$2C(coal)+2H_2O(g)\rightarrow CH_4(g)+CO_2(g)$ $\Delta H_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $C(coal)+H_2O(g)\rightarrow CO(g)+H_2(g)$ $\Delta H_1=29.7kJ$

(2) $CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)$ $\Delta H_2=-41kJ$

(3) $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)$ $\Delta H_3=-206kJ$

We are multiplying equation 1 by 2 and then adding all the equations, we get :

(b) The expression for enthalpy of reaction will be,

$\Delta H_{rxn}=2\times \Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H_{rxn}=(2\times 29.7)+(-41)+(-206)$

$\Delta H_{rxn}=-187.6kJ/mol$

Therefore, the enthalpy of reaction is, -187.6 kJ/mol

(c) Now we have to calculate the total heat.

$\Delta H=\frac{q}{n}$

or,

$q=\Delta H\times n$

where,

$\Delta H$ = enthalpy change = -187.6 kJ/mol

q = heat = ?

n = number of moles of coal = $\frac{1.00\times 1000g}{12.00g/mol}=83.33mol$

Now put all the given values in the above formula, we get:

$q=(-187.6kJ/mol)\times (83.33mol)=-2.251kJ$

Thus, the total heat will be, -2251 kJ

4 0

2 years ago

A different student is given a 10.0g sample labeled CaBr2 that may contain an inert (nonreacting) impurity. Identify a quantity

grandymaker [24]

Answer:

Mass of Ca in sample, Mass of Br in sample, Number of moles of Ca in sample, Number of moles of Br in sample, Mass or moles of element other than Ca or Br in sample

Explanation:

The AP Classroom will not count your answer to this question as correct unless it includes at least one of the answers listed above. If you say that theanswer to this question is density, it will be marked as incorrect, I found that out the hard way when I used the answers that brainly gave me.

Good luck,

I applaud you for using the sources avalible to you, which is /definetly not/ cheeting.

8 0

1 year ago

A vessel contains Ar(g) at a high pressure. Which of the following statements best helps to explain why the measured pressure is

Luda [366]

Answer:

4. The combined volume of the Ar atoms is too large to be negligible compared with the total volume of the container.

Explanation:

Deviations from ideality are due to intermolecular forces and to the nonzero volume of the molecules themselves. At infinite volume, the volume of the molecules themselves is negligible compared with the infinite volume the gas occupies.

However, the volume occupied by the gas molecules must be taken into account. Each molecule does occupy a finite, although small, intrinsic volume.

The non-zero volume of the molecules implies that instead of moving in a given volume V they are limited to doing so in a smaller volume. Thus, the molecules will be closer to each other and repulsive forces will dominate, resulting in greater pressure than the one calculated with the ideal gas law, that means, without considering the volume occupied by the molecules.

5 0

2 years ago

An unopened, cold 2.00 L bottle of soda contains 46.0 mL of gas confined at a pressure of 1.30 atm at a temperature of 5.0oC. If

Katena32 [7]

Answer:

20.8mL

Explanation:

7 0

2 years ago

What pressure (in atm) would be exerted by 76 g of fluorine gas (f2) in a 1.50 liter vessel at -37oc? (a) 26 atm(b) 4.1 atm(c) 1

nirvana33 [79]

Let's assume that the F₂ gas has ideal gas behavior.

Then we can use ideal gas formula,
PV = nRT

Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol⁻¹ K⁻¹) and T is temperature in Kelvin.

Moles = mass / molar mass

Molar mass of F₂ = 38 g/mol

Mass of F₂ = 76 g

Hence, moles of F₂ = 76 g / 38 g/mol = 2 mol

P = ?
V = 1.5 L = 1.5 x 10⁻³ m³

n = 2 mol

R = 8.314 J mol⁻¹ K⁻¹
T = -37 °C = 236 K

By substitution,

P x 1.5 x 10⁻³ m³ = 2 mol x 8.314 J mol⁻¹ K⁻¹ x 236 K

p = 2616138.67 Pa

p = 25.8 atm = 26 atm

Hence, the pressure of the gas is 26 atm.

Answer is "a".

5 0

1 year ago