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3 years ago

what is the chemical equation in words for 2 C2H2 (g) + 5 O2 (g) ---------> 4 CO2 (g) + 2 H2O (l )

Chemistry

1 answer:

elena-s [515]3 years ago

4 0

Answer:

Ethyn(g) + 5oxygen(g) ------------> 4carbondioxide(g) + 2water(l)

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Which of these solutions are basic at 25 Â°C? Solution A: [OHâˆ’]=3.13Ã—10âˆ’7 M Solution C: [H3O+]=0.000747 M Solution B: [H3O+

solong [7]

Answer:

Are basic:

[OH⁻] = 3.13x10⁻⁷M and [H₃O⁺] = 9.55x10⁻⁹M

Explanation:

A solution is basic when pH = - log [H₃O⁺] is higher than 7.

It is possible to convert [OH⁻] to [H₃O⁺] using:

[H₃O⁺] = 1x10⁻¹⁴ / [OH⁻]

a. [OH⁻] = 3.13x10⁻⁷M

[H₃O⁺] = 1x10⁻¹⁴ / [3.13x10⁻⁷M]

[H₃O⁺] = 3.19x10⁻⁸M

pH = - log [H₃O⁺] = 7.50

[OH⁻] = 3.13x10⁻⁷M is basic

b. pH = -log [H₃O⁺] = - log 0.000747M = 3.13.

This solution is not basic

c. [H₃O⁺] = 9.55x10⁻⁹M

pH = 8.02

This solution is also basic.

8 0

2 years ago

Part 1: A cylinder containing 20.0 L of compressed nitrogen is connected to an empty (evacuated) vessel with an unknown volume.

baherus [9]

Answer:

The volume of the vessel is 250 L

Partial pressure of hydrogen = 189 torr

Explanation:

Using Boyle's law

${P_1}\times {V_1}={P_2}\times {V_2}$

Given ,

V₁ = 20.0 L

V₂ = ?

P₁ = 25 atm

P₂ = 2 atm

Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${25}\times {20.0}={2}\times {V_2}$

${V_2}=\frac {{25}\times {20.0}}{2}\ L$

${V_2}=250\ L$

<u>The volume of the vessel is 250 L.</u>

According to Dalton's law of partial pressure:-

$P_{H_2}=Mole\ fraction\times Total\ Pressure$

So, according to definition of mole fraction:

$Mole\ fraction\ of\ H_2=\frac {n_{H_2}}{n_{H_2}+n_{He}}$

Also,

Mole fraction of H₂ = 1 - Mole fraction of He = 1 - 0.75 = 0.25

So,

Total pressure = 756 torr

Thus,

$P_{H_2}=0.25\times 756\ torr$

<u>Partial pressure of hydrogen = 189 torr.</u>

7 0

3 years ago

The rate of effusion of an unknown gas was measured and found to be 11.9 mL/min. Under identical conditions, the rate of effusio

iren2701 [21]

Answer : The correct option is, (B) $CO_2$

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

$R\propto \sqrt{\frac{1}{M}}$

or,

$(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}$ ..........(1)

where,

$R_1$ = rate of effusion of unknown gas = $11.9\text{ mL }min^{-1}$

$R_2$ = rate of effusion of oxygen gas = $14.0\text{ mL }min^{-1}$

$M_1$ = molar mass of unknown gas = ?

$M_2$ = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the above formula 1, we get:

$(\frac{11.9\text{ mL }min^{-1}}{14.0\text{ mL }min^{-1}})=\sqrt{\frac{32g/mole}{M_1}}$

$M_1=44.2g/mole$

The unknown gas could be carbon dioxide $(CO_2)$ that has approximately 44 g/mole of molar mass.

Thus, the unknown gas could be carbon dioxide $(CO_2)$

5 0

3 years ago

In beta decay, what is emitted?

Bingel [31]

I'm pretty sure its an electron. I hope this helps! (:

5 0

3 years ago

Read 2 more answers

PLEASE HELP ME ASAP THANK YOU!

fomenos

Answer:

The activation energy was reached was 10:45 a.m. The additional energy did not affect the reaction.

Explanation:

7 0

3 years ago

Read 2 more answers