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3 years ago

Please help!!!!!!!!!!!!!!!!!!!

Chemistry

1 answer:

zimovet [89]3 years ago

3 0

Answer:

Your body cells use the oxygen you breathe to get energy from the food you eat. This process is called cellular respiration. During cellular respiration the cell uses oxygen to break down sugar. Breaking down sugar produces the energy your body needs.

Explanation:

Brainly pls

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A ground test utilizing an auxiliary current electrode and an auxiliary potential electrode is known as the______.

kompoz [17]

Answer:

The three-point test

Explanation:

The three-point test refers to a ground test utilizing an auxiliary current electrode and an auxiliary potential electrode.

7 0

2 years ago

BTW IT IS SCIENCE

Sliva [168]

The answer is A warm air rises cool air sinks

4 0

3 years ago

Read 2 more answers

When might a large volume of material have little mass?

il63 [147K]

For a solid I would say something like aerogel would have a large volume but very little mass.

Let me know if this helped or not.

5 0

4 years ago

Using the standard enthalpies of formation for the chemicals involved, calculate the enthalpy change for the following reaction.

Firlakuza [10]

Answer: -134 kJ

Explanation:

The balanced chemical reaction is,

$3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq)+NO(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

$\Delta H=[(n_{HNO_3}\times \Delta H_{HNO_3})+(n_{NO}\times \Delta H_{NO})]-[(n_{H_2O}\times \Delta H_{H_2O})+(n_{NO_2}\times \Delta H_{NO_2})]$

where,

n = number of moles

Now put all the given values in this expression, we get

$\Delta H=[(2\times -207)+(1\times 90)]-[(1\times -286)+(3\times 32)]$

$\Delta H=-134kJ$

Therefore, the enthalpy change for this reaction is, -134 kJ

6 0

3 years ago

At 14,000 ft elevation the air pressure drops to 0.59 atm. Assume you take a 1L sample of air at this altitude and compare it to

Ray Of Light [21]

Answer:

There are 0.1125 g of O₂ less in 1 L of air at 14,000 ft than in 1 L of air at sea level.

Explanation:

To solve this problem we use the ideal gas law:

PV=nRT

Where P is pressure (in atm), V is volume (in L), n is the number of moles, T is temperature (in K), and R is a constant (0.082 atm·L·mol⁻¹·K⁻¹)

Now we calculate the number of moles of air in 1 L at sea level (this means with P=1atm):

1 atm * 1 L = n₁ * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 K

n₁=0.04092 moles

Now we calculate n₂, the number of moles of air in L at an 14,000 ft elevation, this means with P = 0.59 atm:

0.59 atm * 1 L = n₂ * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 K

n₂=0.02414 moles

In order to calculate the difference in O₂, we substract n₂ from n₁:

0.04092 mol - 0.02414 mol = 0.01678 mol

Keep in mind that these 0.01678 moles are of air, which means that we have to look up in literature the content of O₂ in air (20.95%), and then use the molecular weight to calculate the grams of O₂ in 20.95% of 0.01678 moles:

$0.01678mol*\frac{20.95}{100} *32\frac{g}{mol} =0.1125 gO_{2}$

4 0

4 years ago