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kolbaska11 [484]

3 years ago

10

Suppose there is 0.63 g of HNO3 per 100 mL of a particular solution. What is the concentration of the HNO3 solution in moles per

liter?

1 answer:

Vadim26 [7]3 years ago

4 0

Answer:

There are 0.09996826 moles per liter of the solution.

Explanation:

Molar mass of HNO3: 63.02

Convert grams to moles

0.63 grams/ 63.02= 0.009996826

Convert mL to L and place under moles (mol/L)

100mL=0.1 L

0.009996826/0.1= 0.09996826 mol/L

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If a gas is cooled from 325K to 275K and the volume is constant, what would the final pressure be if it was originally 750mmHg?

salantis [7]

Answer:

635

p/T is a constant

(750/325)×275

3 0

4 years ago

Calculate the volume of 0,004 moles of H2 at STP <br>

Angelina_Jolie [31]

Answer:

Explanation:

At STP, 1 mole of hydrogen can occupy a volume of 22.4 liters. At STP, 3 moles of hydrogen can occupy a volume of 2×22. 4=44. 8 liters

7 0

2 years ago

A sample of ammonia ^NH3h gas is completely decomposed to nitrogen and hydrogen gases over heated iron wool. If the total pressu

icang [17]

Answer : The partial pressure of $N_2$ and $H_2$ is, 216.5 mmHg and 649.5 mmHg

Explanation :

According to the Dalton's Law, the partial pressure exerted by component 'i' in a gas mixture is equal to the product of the mole fraction of the component and the total pressure.

Formula used :

$p_i=X_i\times p_T$

$X_i=\frac{n_i}{n_T}$

So,

$p_i=\frac{n_i}{n_T}\times p_T$

where,

$p_i$ = partial pressure of gas

$X_i$ = mole fraction of gas

$p_T$ = total pressure of gas

$n_i$ = moles of gas

$n_T$ = total moles of gas

The balanced decomposition of ammonia reaction will be:

$2NH_3\rightarrow N_2+3H_2$

Now we have to determine the partial pressure of $N_2$ and $H_2$

$p_{N_2}=\frac{n_{N_2}}{n_T}\times p_T$

Given:

$n_{N_2}=1\\\\n_{H_2}=3\\\\n_{T}=4\\\\p_T=866mmHg$

$p_{N_2}=\frac{1}{4}\times (866mmHg)=216.5mmHg$

and,

$p_{H_2}=\frac{n_{H_2}}{n_T}\times p_T$

Given:

$n_{H_2}=1\\\\n_{H_2}=3\\\\n_{T}=4\\\\p_T=866mmHg$

$p_{H_2}=\frac{3}{4}\times (866mmHg)=649.5mmHg$

Thus, the partial pressure of $N_2$ and $H_2$ is, 216.5 mmHg and 649.5 mmHg

5 0

4 years ago

The molar volume of a certain solid is 142.0 cm3 mol−1 at 1.00 atm and 427.15 k, its melting temperature. the molar volume of th

kotykmax [81]

Answer:

$\Delta _{fus}H=3255.3J/mol$

$\Delta _{fus}S=7.62\frac{J}{mol*K}$

Explanation:

Hello,

Clausius Clapeyron equation is suitable in this case, since it allows us to relate the P,T,V behavior along the described melting process and the associated energy change. Such equation is:

$\frac{dp}{dT}=\frac{\Delta _{fus}H}{T\Delta _{fus}V}$

As both the enthalpy and volume do not change with neither the temperature nor the pressure for melting processes, its integration turns out:

$p_2-p_1=\frac{\Delta _{fus}H}{\Delta _{fus}V}ln(\frac{T_2}{T_1} )$

Solving for the enthalpy of fusion we obtain:

$\Delta _{fus}H=\frac{(p_2-p_1)(V_2-V1)}{ln(\frac{T_2}{T_1})} =\frac{(11.84atm-1.00 atm)(156.6cm^3/mol-142.0cm^3/mol)}{ln(\frac{429.26K}{427.15K} )} \\\\\Delta _{fus}H=32127.3atm*cm^3/mol*\frac{101325Pa}{1atm}*(\frac{1m}{100cm} )^3\\\Delta _{fus}H=3255.3J/mol$

Finally the entropy of fusion is given by:

$\Delta _{fus}S=\frac{\Delta _{fus}H}{T_1} =\frac{3255.3J/mol}{427.15K}\\ \\\Delta _{fus}S=7.62\frac{J}{mol*K}$

Best regards.

5 0

3 years ago

A molecule contains four bonded pairs of electrons and zero lone pairs. What is the name of the molecular geometry?

vagabundo [1.1K]

Answer: Option (d) is the correct answer.

Explanation:

A tetarhedral geometry is the geometry which includes four atoms bonded to the central atom and it does not contain any lone pair of electron.

For example, $CH_{4}$ has tetrahedral geometry.

Thus, we can conclude that a molecule which contains four bonded pairs of electrons and zero lone pairs. The name of the molecular geometry is tetrahedral.

4 0

3 years ago