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2 years ago

How many molecules are in 2.50 moles of co2

Chemistry

2 answers:

lapo4ka [179]2 years ago

6 0

Explanation:

There are 1.51 x 1024 molecules of carbon dioxide in 2.50 moles of carbon dioxide.

zmey [24]2 years ago

4 0

Answer:

$1.51 \times 1024 \: \: molecules$

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1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.

Andrej [43]

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

$Mg+2HCl\rightarrow MgCl_2+H_2$

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

$n_{H_2}^{by\ HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl} =0.0284molH_2\\\\n_{H_2}^{by\ Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2$

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

$m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2$

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

$m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg} =0.690gMg$

Thus, the mass of excess magnesium turns out:

$m_{Mg}^{excess}=2.38g-0.690g=1.69gMg$

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

$Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%$

Best regards!

8 0

2 years ago

A student has a 2.19 L bottle that contains a mixture of O 2 , N 2 , and CO 2 with a total pressure of 5.57 bar at 298 K . She k

Sergeeva-Olga [200]

<u>Answer:</u> The partial pressure of oxygen gas is 2.76 bar

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 5.57 bar

V = Volume of the gas = 2.19 L

T = Temperature of the gas = 298 K

R = Gas constant = $0.0831\text{ L bar }mol^{-1}K^{-1}$

n = Total number of moles = ?

Putting values in above equation, we get:

$5.57bar\times 2.19L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{5.57\times 2.19}{0.0831\times 298}=0.493mol$

To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

$p_{A}=p_T\times \chi_{A}$ ........(1)

where,

$p_A$ = partial pressure of carbon dioxide = 0.318 bar

$p_T$ = total pressure = 5.57 bar

$\chi_A$ = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

$0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas, $\chi_{N_2}=\frac{0.221}{0.493}=0.448$

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

$p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar$

Hence, the partial pressure of oxygen gas is 2.76 bar

6 0

3 years ago

Which statement best describes what makes a base weak?

dimulka [17.4K]

A base is weak when only a little of it dissolved in Water.

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3 years ago

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