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qwelly [4]
3 years ago
8

Help me if you can please help me please​

Chemistry
2 answers:
Dovator [93]3 years ago
8 0

Answer: with? i can help:))

Explanation:

hmu love

olga2289 [7]3 years ago
6 0

Answer:

what you need help with ??

Explanation:

Il help

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Answer:

yo try and search it on Google

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If a sodium atom looses an electron to form an ion what is its charge?
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Difference between giant ionic structure,giant covalent structure and simple molecular structure and also covalent bonds and ion
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. How many grams of Na2SO4 are required to make 2700 mL of a 2.0 M solution?
ratelena [41]

Answer:

Explanation:

From the net ionic equation

Ba2+(aq) + SO42-(aq) ==> BaSO4(s) we see that 1 mole Ba2+ reacts with 1 mole SO42- to -> 1 mol BaSO4

Find moles of Ba2+ used: 0.250 moles/L x 0.0323 L = 0.008075 moles Ba2+

Find moles SO42- present: 0.008075 moles Ba2+ x 1 mol SO42-/1 mol Ba2+ = 0.008075 mol SO42-

Find mass of Na2SO4 present: 0.008075 mol SO42- x 1 mol Na2SO4/1 mol SO42- x 142.04 Na2SO4/mole = 1.14698 g = 1.15 g Na2SO4 (to 3 significant figures)

6 0
2 years ago
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5. What are the relative rates of diffusion for methane, CH, and oxygen, O2? If O2 la travels 1.00 m in a certain amount of time
11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

We are given:

Molar mass of methane gas, m = 16 g/mol

Molar mass of oxygen gas,m' = 32 g/mol

By taking their ratio, we get:

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{m'}{m}}

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{32}{16}}=1.4142

The relative rates of diffusion for methane and oxygen is 1.4142.

If oxygen gas travels 1 meters in time t.

Rate of diffusion of oxygen =d_{O_2}=\frac{1 m}{t}

If methane gas travels travels in y meters in time t.

Rate of diffusion of methane=d_{CH_4}=\frac{y }{t}

\frac{d_{CH_4}}{d_{O_2}}=\frac{\frac{y }{t}}{\frac{1 m}{t}}=1.4142

y = 1.4142 m

Methane gas will be able to travel 1.4142 meter in the same conditions.

8 0
3 years ago
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