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schepotkina [342]

2 years ago

10

Question 1 how many moles of helium are found in a balloon that contains 5.5 l of helium at a pressure of 1.15 atm and a tempera

ture of 22.0 ºc?

2 answers:

DaniilM [7]2 years ago

7 0

Answer : The number of moles of helium gas found in a balloon are 0.26 moles.

Explanation :

To calculate the moles of helium gas, we use the equation given by ideal gas :

PV = nRT

where,

P = Pressure of helium gas = 1.15 atm

V = Volume of the helium gas = 5.5 L

n = number of moles of helium gas = ?

R = Gas constant = $0.0821\text{ L.atm }mol^{-1}K^{-1}$

T = Temperature of helium gas = $22.0^oC=273+22.0=295K$

Putting values in above equation, we get:

$1.15atm\times 5.5L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 295K\\\\n=0.26mol$

Therefore, the number of moles of helium gas found in a balloon are 0.26 moles.

denpristay [2]2 years ago

3 0

Using pv=nRT
T= 22+273 K
P=1.15 Pa
R=8.31
V =5.5... (u did not write the unit so...)

Therefore n,mole= (1.15×5.5) ÷ (273+22)(8.31)

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A 2.550 x 10^−2 M solution of glycerol (C3H8O3) in water is at 20.0°C. The sample was created by dissolving a sample of C3H8O3 i

BARSIC [14]

Answer:

$2.557\times 10^{-2}\ molal.$

Explanation:

Given , molarity of glycerol= $2.55\times 10^-^2\ M.$

Volume= 1 L.

Therefore, No of moles of glycerol= $molarity\times volume\ in\ liters=2.55\times 10^-^2\ moles.$

Now, volume of water needed, V=998.8 mL.

Density is given as= 0.9982 g/mL.

Therefore, mass of water = $volume\times density=998.8\times 0.9982=997\ g=0.997\ kg$

Now, molality= $\dfrac{no\ of\ moles}{mass\ of\ solvent\ in\ kg}=\dfrac{2.55\times 10^-^2\ }{0.997}=0.02557\ molal=2.557\times 10^{-2}\ molal.$

Hence, this is the required solution.

3 0

1 year ago

How many moles of lead (ii) chromate are in 51 grams of this substance? answer in units of mol?

maria [59]

Mass of lead (II) chromate is 51 g. The molecular formula is $PbCrO_{4}$ and its molar mass is 323.2 g/mol

Number of moles can be calculated using the following formula:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Putting the values,

$n=\frac{(51 g}{323.1937 g/mol}=0.1578 mol$

Therefore, number of moles of lead (II) chromate will be 0.1578 mol.

5 0

2 years ago

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

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Answer:

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enot [183]

It can be made true by changing "cannot" to "can".

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2 years ago