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3 years ago

For a half-reaction in an acidic solution, which substances should you add to balance the equation?

1 answer:

8 0

The answer is a. H+ and H2O. For a half-reaction in acidic solutions, elements that are not hydrogen and oxygen are balanced first. Then we add H2O to balance oxygen atoms and we add H+ to balance the hydrogen atoms. We do the same steps for basic solutions, but we also add hydroxide ions OH- in the latter part.

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What is the percent composition of tungsten

lawyer [7]

20 is the answer to your question

5 0

4 years ago

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How many grams of CO(g) are there in 74.5 mL of the gas at 0.933 atm and 30o C?

Helen [10]

Answer:

As per the given statement:

$V = 74.5mL$

$P = 0.933 atm$

$T = 30^{\circ} C$

Using ideal gas law equation:

$PV = nRT$

where

P represents the Pressure of a gas

V represents the Volume it occupies

R represents the Universal constant , usually given as:

R = 0.082 atm L/ mol k

First convert Celsius into Kelvin.

T = 273 + 30 = 303 K

Use conversion:

1 L = 1000 mL

V = 74.5 mL = $74.5 \times 10^{-3}$ L

Now substitute the given values we have;

$n = \frac{PV}{RT}$

$n = \frac{0.933 \cdot 74.5 \cdot 10^{-3}}{0.082 \cdot 303}$

Simplify:

n = 0.00279757305

n ≈0.0028 moles.

Since carbon monoxide, CO has molar mass of 28.01 g/mol

then;

$m = 0.0028 \times 28.01 = 0.078$ g

Therefore, 0.078 g of CO(g) are there in 74.5 mL of the gas at 0.933 atm and 30 degree C

4 0

3 years ago

True or false: The number of protons an element has can vary.

Lerok [7]

Answer:

true

Explanation:

5 0

3 years ago

Read 2 more answers

When the vapor pressure is less than the atmospheric pressure, what will occur? A. The liquid will not boil. B. the liquid will

AfilCa [17]

Answer:

A. the liquid will not boil

5 0

3 years ago

A sample of a gas occupies 2.76 L at 303K. What would the volume be if the temperature was increased to 404K? Assume constant

Artist 52 [7]

Answer:

If the temperature was increased to 404 K, its volume would be 3.68 L.

Explanation:

Charles' Law gives a relationship between the volume and the temperature of the gas at constant temperature. This law states that the volume of a given amount of gas held at constant pressure is directly proportional to the temperature.

$V\propto T$

$\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}$

Let

$V_1=2.76\ L\\\\T_1=303\ K\\\\T_2=404\ K$

Let $V_2$ is new volume. Using above formula we get :

$V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{2.76\times 404}{303}\\\\V_2=3.68\ L$

If the temperature was increased to 404 K, its volume would be 3.68 L.

6 0

4 years ago