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grandymaker [24]
4 years ago
7

Which do you think could work to reduce air pollution and global warming

Chemistry
1 answer:
natka813 [3]4 years ago
5 0
Recycling!! And also planting more trees.

Hope this helps!
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Henok is observing the reaction rate of a chemical reaction. He then adds a catalyst to the reaction.
tia_tia [17]
A catalyst will ALWAYS increase the reaction rate so C
4 0
3 years ago
.
Natasha_Volkova [10]

Answer:

(C) through the atmosphere

Explanation:

6 0
3 years ago
Read 2 more answers
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
Can you please help me and can you show your work please
natali 33 [55]

Answer:

1. 25 moles water.

2. 41.2 grams of sodium hydroxide.

3. 0.25 grams of sugar.

4. 340.6 grams of ammonia.

5. 4.5x10²³ molecules of sulfur dioxide.

Explanation:

Hello!

In this case, since the mole-mass-particles relationships are studied by considering the Avogadro's number for the formula units and the molar mass for the mass of one mole of substance, we proceed as shown below:

1. Here, we use the Avogadro's number to obtain the moles in the given molecules of water:

1.5x10^{25}molecules*\frac{1mol}{6.022x10^{23}molecules} =25 molH_2O

2. Here, since the molar mass of NaOH is 40.00 g/mol, we obtain:

1.2mol*\frac{40.00g}{1mol} =41.2g

3. Here, since the molar mass of C6H12O6 is 180.15 g/mol:

45g*\frac{1mol}{180.15g}=0.25g

4. Here, since the molar mass of ammonia is 17.03 g/mol:

20mol*\frac{17.03g}{1mol}=340.6g

5. Here, since the molar mass of SO2 is 64.06 g/mol:

48g*\frac{1mol}{64.06g} *\frac{6.022x10^{23}molecules}{1mol} =4.5x10^{23}molecules

Best regards!

5 0
3 years ago
In testing the effectiveness of an antacid compound, 20.0g of Hydrochloric Acid is mixed with 28.0g of Magnesium Hydroxide. Will
BartSMP [9]

Answer:

Base Mg(OH)2 does neutralise the acid and is 12g in excess.

Explanation:

2HCL +Mg(OH)2 -> MgCl2 + 2H20

2 * 36.458 g of HCL react with  58.319 g of  Mg(OH)2 to neutralise it.

72.916 HCl reacts with  58.319 g of the base.

So 20 g HCl reacts with  (58.319/72.916) * 20 = 16g.

There are 28 g of Mg(OH)2 so the base does neutralise all the acid.

The Mg(OH)2 is 28 - 16 = 12 g in excess.

4 0
3 years ago
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