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2 years ago

Could someone please help me with this chemistry question I will mark the correct answer as brainliest

Chemistry

1 answer:

Arisa [49]2 years ago

8 0

It is 95% ethanol and 5%water
I’m pretty sure hope you get it right!
:)

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A gaseous mixture composed of 20% CH4, 30% C2H4, 35% C2H2, and 15% C2H20. What is the average molecular weight of the mixture? a

Semenov [28]

<u>Answer:</u> The correct answer is Option d.

<u>Explanation:</u>

We are given:

Mass percentage of $CH_4$ = 20 %

So, mole fraction of $CH_4$ = 0.2

Mass percentage of $C_2H_4$ = 30 %

So, mole fraction of $C_2H_4$ = 0.3

Mass percentage of $C_2H_2$ = 35 %

So, mole fraction of $C_2H_2$ = 0.35

Mass percentage of $C_2H_2O$ = 15 %

So, mole fraction of $C_2H_2O$ = 0.15

We know that:

Molar mass of $CH_4$ = 16 g/mol

Molar mass of $C_2H_4$ = 28 g/mol

Molar mass of $C_2H_2$ = 26 g/mol

Molar mass of $C_2H_2O$ = 48 g/mol

To calculate the average molecular mass of the mixture, we use the equation:

$\text{Average molecular weight of mixture}=\frac{_{i=1}^n\sum{\chi_im_i}}{n_i}$

where,

$\chi_i$ = mole fractions of i-th species

$m_i$ = molar masses of i-th species

$n_i$ = number of observations

Putting values in above equation:

$\text{Average molecular weight}=\frac{(\chi_{CH_4}\times M_{CH_4})+(\chi_{C_2H_4}\times M_{C_2H_4})+(\chi_{C_2H_2}\times M_{C_2H_2})+(\chi_{C_2H_2O}\times M_{C_2H_2O})}{4}$

$\text{Average molecular weight of mixture}=\frac{(0.20\times 16)+(0.30\times 28)+(0.35\times 26)+(0.15\times 42)}{4}\\\\\text{Average molecular weight of mixture}=6.75$

Hence, the correct answer is Option d.

3 0

3 years ago

Increasing the temperature, as noted in the graph, increases the rate of this chemical reaction. What is the relationship betwee

Soloha48 [4]

Answer: (C) Although the average kinetic energy of the colliding substances increases, this has no influence on activation energy.

Explanation:

After increasing the temperature of the reaction , the rate of the chemical reaction increases due to increase in the average kinetic energy of the particles. At increased temperature high proportions of particles can react making the reaction faster.

8 0

3 years ago

Which of these zones of the ocean is most hospitable to life?

Nat2105 [25]

The answer is epipelagic zone

3 0

2 years ago

Read 2 more answers

What is the change in internal energy ( ΔU ) of the system if q = –8 kJ and w = –1 kJ for a certain process?

Radda [10]

Answer:

Change in internal energy (ΔU) = -9 KJ

Explanation:

Given:

q = –8 kJ [Heat removed]

w = –1 kJ [Work done]

Find:

Change in internal energy (ΔU)

Computation:

Change in internal energy (ΔU) = q + w

Change in internal energy (ΔU) = -8 KJ + (-1 KJ)

Change in internal energy (ΔU) = -8 KJ - 1 KJ

Change in internal energy (ΔU) = -9 KJ

6 0

3 years ago

20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

3 0

3 years ago