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2 years ago

Choose all the answers that apply.

Chemistry

1 answer:

Anni [7]2 years ago

3 0

Answer:

I believe it's the lowest portion of the atmosphere

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Calculate the energy E of a sample of 3.50 mol of ideal oxygen gas (O2) molecules at a temperature of 310 K. Assume that the mol

love history [14]

Answer:

13.53 kJ

Explanation:

The energy of a gas can be calculated by the equation:

E = (3/2)*n*R*T

Where n is the number of moles, R is the gas constant (8.314 J/mol.K), and T is the temperature.

E = (3/2)*3.5*8.314*310

E = 13,531.035 J

E = 13.53 kJ

7 0

3 years ago

Which of the answer choices best describes elements?

lianna [129]

It can be a compound or a single element. An element is a pure substance that cannot be separated into simpler substances by chemical or physical means. There are about 117 elements, butcarbon, hydrogen, nitrogen and oxygen are only a few that make up the largest portion of Earth.

I would say d.

4 0

3 years ago

Using the molarity equation, calculate how many grams of salt you need in order to make 80ml of a 2M solution (MW = 58.44g/mole)

ollegr [7]

Answer:

9.35g

Explanation:

The molarity equation establishes that:

$\textrm{molarity}=\frac{\textrm{moles of solute}}{\textrm{liters of solution}}$

So, we have information about molarity (2M) and volume (80 ml=0.08 l), with that, we can find the moles of solute:

$\textrm{moles of solute}=\textrm{molarity}*\textrm{liters of solution}$

$\textrm{moles of solute}= 0.08 \textrm{ l} *2\textrm{ M} = 0.16 \textrm{ mol}$

The mathematical equation that establishes the relationship between molar weight, mass and moles is:

$\textrm{molar weight}= \frac{\textrm{mass}}{\textrm{moles}}$

$\textrm{MW}= \frac{\textrm{m}}{\textrm{n}}$

We have MW (58.44g/mole) and n (0.16 mol), and we need to find m (grams of salt needed) to solve the problem:

$\textrm{m} = \textrm{MW * n}= 58.44\frac{\textrm{g}}{\textrm{mol}} * 0.16 \textrm{ mol} = 9.35 \textrm{ g}$

8 0

3 years ago

How would I do this? A beaker contains 50.0 mL of 0.25 M aluminum nitrate solution. What is the minimum volume of 0.2 M sodium s

Serhud [2]

The minimum volume of 0.2M sodium sulfide that will precipitate out aluminum from 50.0 mL, 0.25 M aluminum nitrate would be 0.094 L or 94 mL

<h3>Stoichiometric calculation</h3>

From the equation of the reaction:

$2Al(NO_3)_3 + 3Na_2S ---> Al_2S_3 + 6NaNO_3$

Mole ratio of Na2S and Al(NO3)3 = 3:2

Mole of 50.0 mL, 0.25 M Al(NO3)3 = 50/1000 x 0.25

= 0.0125 mole

Equivalent mole of Na2S = 3/2 x 0.0125

= 0.0188 mole

Volume of 0.2M, 0.0188 mole Na2S = 0.0188/0.2

= 0.094 L or 94 mL

More on stoichiometric calculations can be found here: brainly.com/question/8062886

6 0

2 years ago

Calculate the mass in grams of 4.69×1024 molecules of methanol.

Vladimir [108]

Answer:

4802.56

Explanation:

5 0

2 years ago