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3 years ago

What is the molar concentration of 5% acetic acid?

Chemistry

1 answer:

larisa [96]3 years ago

3 0

Answer:Vinegar Is 5% Acetic Acid And Its Molarity Is 0.833M.

Explanation:

yes

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Name the fundamental unit involved in the derived unit joule?

tangare [24]

Answer:

Energy

Explanation:

7 0

2 years ago

For the following reaction, C6H12O6 and O2 are the

erastovalidia [21]

Thats burning of glucose as fuel in the cells. It generates heat and is an exothermic reaction. In Mitochondria, which act as furnaces, the fuel helps generate heat for the cells of the body.

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3 years ago

2. How many moles are in 8.30 x 1023 molecules of H2O?

daser333 [38]

Answer:

<h2>1.38 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{8.30 \times {10}^{23} }{6.02 \times {10}^{23} } = \frac{8.30}{6.02} \\ = 1.3787 37...$

We have the final answer as

<h3>1.38 moles</h3>

Hope this helps you

8 0

3 years ago

Carbon monoxide (CO) reacts with hydrogen (H2) to form methane (CH4) and water (H2O).

Artemon [7]

Answer:

5.9x10^-2 M

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Concentration of CO, [CO] = 0.30 M

Concentration of H2, [H2] = 0.10 M

Concentration of H2O, [H2O] = 0.020 M

Equilibrium constant, K = 3.90

Concentration of CH4, [CH4] =..?

Step 2:

The balanced equation for the reaction. This is given below:

CO(g) + 3H2(g) <=> CH4(g) + H2O(g)

Step 3:

Determination of the concentration of CH4.

The expression for equilibrium constant of the above equation is given below:

K = [CH4] [H2O] / [CO] [H2]^3

3.9 = [CH4] x 0.02/ 0.3 x (0.1)^3

Cross multiply to express in linear form

[CH4] x 0.02= 3.9 x 0.3 x (0.1)^3

Divide both side by 0.02

[CH4] = 3.9 x 0.3 x (0.1)^3 /0.02

[CH4] = 5.9x10^-2 M

Therefore, the equilibrium concentration of CH4 is 5.9x10^-2 M

5 0

2 years ago

The water in a pressure cooker boils at a temperature greater than 100°C because it is under pressure. At this higher temperatur

vazorg [7]

Answer:

the activation energy Ea = 179.176 kJ/mol

it will take 7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

Explanation:

From the given information

$T_1 = 100^0 C = 100+273 = 373 \ K \\ \\ T_2 = 113^0 C = 113 + 273 = 386 \ K$

$R_1 = \dfrac{1}{7}$

$R_2 = \dfrac{1}{49}$

Thus; $\dfrac{R_2}{R_1} = 7$

Because at 113.0°C; the rate is 7 time higher than at 100°C

Hence:

$In (7) = \dfrac{Ea}{8.314}( \dfrac{1}{373}- \dfrac{1}{386})$

1.9459 = $\dfrac{Ea}{8.314}* 9.0292 *10^{-5}$

$1.9459*8.314 = Ea * 9.0292*10^{-5}$

$16.1782126= Ea * 9.0292*10^{-5}$

$Ea = \dfrac{16.1782126}{ 9.0292*10^{-5}}$

Ea = 179.176 kJ/mol

Thus; the activation energy Ea = 179.176 kJ/mol

here;

$T_2 = 386 \ K \\ \\T_1 = (89.8 + 273)K = 362.8 \ K$

$In(\dfrac{R_2}{R_1})= \dfrac{Ea}{R}(\dfrac{1}{T_1}- \dfrac{1}{T_2})$

$In(\dfrac{R_2}{R_1})= \dfrac{179.176}{8.314}(\dfrac{1}{362.8}- \dfrac{1}{386})$

$In (\dfrac{R_2}{R_1}) = 0.00357$

$\dfrac{R_2}{R_1}= e^{0.00357}$

$\dfrac{R_2}{R_1}= 1.0035$

where ;

$R_2 = \dfrac{1}7{}$

$R_1 = \dfrac{1}{t}$

Now;

$\dfrac{t}{7}= 1.0035$

t = 7.0245 mins

Therefore; it will take 7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

4 0

3 years ago