Ask question

Ask question

All categories

3 years ago

6

The atmosphere is a(n) _____. 1. element 2. compound 3. mixture

2 answers:

prohojiy [21]3 years ago

8 0

The atmosphere is a layer of gasses surrounding the planet.
Our atmosphere is a mixture of elements like nitrogen and oxygen, and compounds like carbon dioxide and water.

tensa zangetsu [6.8K]3 years ago

5 0

The atmosphere is a mixture of gasses.

You might be interested in

If a rock had a mass of 20g explain how you would fund the density

REY [17]

Answer:

you have to divide it

Explanation:

4 0

3 years ago

Read 2 more answers

How do alpha particles compare to gamma rays?

Fofino [41]

Answer:

Unlike alpha and beta particles, which have both energy and mass, gamma rays are pure energy. Gamma rays are similar to visible light, but have much higher energy. Gamma rays are often emitted along with alpha or beta particles during radioactive decay.

-Radiation basics.

Explanation:

5 0

4 years ago

Read 2 more answers

How do I calculate the pH of an 11mL sample of C8H4O4 (a mix of NaOH and KHP) when the ka=1.5x10^-5?

Len [333]

Answer:

ummm 3?.................

7 0

3 years ago

If 588 grams of FeS2 is allowed to react with 352 grams of O2 according to the following equation, how many grams of Fe2O3 are p

Rainbow [258]

<u>Answer:</u> The mass of iron (III) oxide produced is 782.5 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For </u> $FeS_2$ <u> :</u>

Given mass of $FeS_2$ = 588 g

Molar mass of $FeS_2$ = 120 g/mol

Putting values in equation 1, we get:

$\text{Moles of }FeS_2=\frac{588g}{120g/mol}=4.9mol$

<u>For </u> $O_2$ <u> :</u>

Given mass of $O_2$ = 352 g

Molar mass of $O_2$ = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of }O_2=\frac{352g}{32g/mol}=11mol$

The chemical equation for the reaction of $FeS_2$ and oxygen gas follows:

$FeS_2+O_2\rightarrow Fe_2O_3+SO_2$

By Stoichiometry of the reaction:

1 mole of $FeS_2$ reacts with 1 mole of oxygen gas

So, 4.9 moles of $FeS_2$ will react with = $\frac{1}{1}\times 4.9=4.9mol$ of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, $FeS_2$ is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of $FeS_2$ produces 1 mole of iron (III) oxide

So, 4.9 moles of $FeS_2$ will produce = $\frac{1}{1}\times 4.9=4.9moles$ of iron (III) oxide

Now, calculating the mass of iron (III) oxide from equation 1, we get:

Molar mass of iron (III) oxide = 159.7 g/mol

Moles of iron (III) oxide = 4.9 moles

Putting values in equation 1, we get:

$4.9mol=\frac{\text{Mass of iron (III) oxide}}{159.7g/mol}\\\\\text{Mass of iron (III) oxide}=(4.9mol\times 159.7g/mol)=782.5g$

Hence, the mass of iron (III) oxide produced is 782.5 grams

8 0

3 years ago

Determine the number of protons, neutrons, and electrons in the following isotopes that are used in medical diagnoses: (a) atomi

il63 [147K]

Answer:

7-c=760

Explanation:

the answer is 7-c=7698

7 0

4 years ago