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2 years ago

Cxdxcfdcxfcf

Chemistry

1 answer:

denis-greek [22]2 years ago

6 0

I honestly don’t know

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You add 7.8 g of iron to 20.70 mL of water and observe that the volume of iron and water together is 21.69 mL . Calculate the de

bixtya [17]

21.69mL - 20.70mL = .99mL Fe
7.8 g / .99 mL = 7.9g/mL

4 0

2 years ago

What is the volume of 1.4 mol of nitrogen gas at STP?

zaharov [31]

Answer:

31.36 Liters

Explanation:

1 mole is equal to 22.4 liters at STP so you can use the equation

1.4 moles * 22.4 liters

to find the volume.

6 0

2 years ago

Calculate the Molarity when a 6.11 mL solution of 0.1 H2SO4 is diluted with 105.12 mL of water

barxatty [35]

Molarity after dilution : 0.0058 M

<h3>Further explanation </h3>

The number of moles before and after dilution is the same

The dilution formula

M₁V₁=M₂V₂

M₁ = Molarity of the solution before dilution

V₁ = volume of the solution before dilution

M₂ = Molarity of the solution after dilution

V₂ = Molarity volume of the solution after dilution

M₁=0.1 M

V₁=6.11

V₂=105.12

$\tt M_2=\dfrac{M_1.V_1}{V_2}=\dfrac{0.1\times 6.11}{105.12}=0.0058~M$

5 0

2 years ago

I need help with the one with the pen on it

Setler [38]

<span>The human body is organized at different levels, starting with the cell. Cells are organized into tissues, and tissues form organs. Organs are organized into organ systems such as the skeletal and muscular systems.</span>

5 0

3 years ago

Read 2 more answers

A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h

frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]$

Where;

$\frac{f_2}{f_1}$ is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1} = 10^{0.7289}\\\\\frac{f_2}{f_1} = 5.356$

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356

5 0

2 years ago