The 225 g FeCl2 is about 1.8 moles.

When 8.00x10^22 molecules of ammonia react with 7.00x10^22 molecules of oxygen according to the chemical equation shown below, h

Pavel [41]

NH₃:

N = 8*10²²
NA = 6.02*10²³

n = N/NA = 8*10²²/6.02*10²³ ≈ 1.33*10⁻¹=0.133mol

O₂:

N=7*10²²
NA = 6.02*10²³

n = N/NA = 7*10²²/6.02*10²³ = 1.16*10⁻¹=0.116mol

4NH₃ <span>+ 3O</span>₂ ⇒<span> 2N</span>₂<span> + 6H</span>₂<span>O
</span>4mol : 3mol : 2mol
0.133mol : 0.116mol : 0,0665mol
limiting reactant

N₂:

n = 0.0665mol
M = 28g/mol

m = n*M = 0.0665mol*28g/mol = <u>1,862g</u>

6 0

3 years ago

The Resource Conservation and Recovery Act controls hazardous waste from its creation to its disposal. Please select the best an

o-na [289]

Answer:

True

Explanation:

Resource Conservation and Recovery Act (RCRA). 1976 of United States Environmental Protection Agency(EPA) empowers EPA to control the production, transportation, storage, treatment and disposal of hazardous waste. The RCRA act was amended in 1984 and 1986 to include Waste minimization along with appropriate disposal (not in the landfill site) and tackling of petroleum hazardous waste respectively along with other waste.

7 0

2 years ago

Read 2 more answers

3 points

VLD [36.1K]

Answer:

<h3>The answer is 8.29 %</h3>

Explanation:

The percentage error of a certain measurement can be found by using the formula

$P(\%) = \frac{error}{actual \: \: number} \times 100\% \\$

From the question

actual density = 19.30g/L

error = 20.9 - 19.3 = 1.6

We have

$p(\%) = \frac{1.6}{19.3} \times 100 \\ = 8.290155440...$

We have the final answer as

Hope this helps you

5 0

3 years ago

How much heat is required to warm 1.50L of water from 25.0C to 100.0C? (Assume a density of 1.0g/mL for the water.)

Masteriza [31]

<u>Answer:</u> The amount of heat required to warm given amount of water is 470.9 kJ

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 1.50 L = 1500 mL (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{1500mL}\\\\\text{Mass of water}=(1g/mL\times 1500mL)=1500g$

To calculate the heat absorbed by the water, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed

m = mass of water = 1500 g

c = heat capacity of water = 4.186 J/g°C

$\Delta T$ = change in temperature = $T_2-T_1=(100-25)^oC=75^oC$

Putting values in above equation, we get:

$q=1500g\times 4.186J/g^oC\times 75^oC=470925J=470.9kJ$

Hence, the amount of heat required to warm given amount of water is 470.9 kJ

6 0

2 years ago

Explain in words or using your diagrams how beryllium atoms would react with fluorine atoms.

Marysya12 [62]

Explanation:

To delineate the the nature of the bonds that would be formed between the two elements, let us first write the electronic configuration of the two species;

Be = 2, 2

F = 2, 7

Beryllium is a metal with two valence electrons whereas fluorine is a halogen with seven valence electrons.

When Be loses two electrons it becomes isoelectronic with He;

Be → Be²⁺ + 2e⁻

Also, when fluorine gains an electron, it becomes isoelectronic with Ne;

F + e⁻ → F⁻

This loss and gain of electrons between the two elements creates an electrostatic attraction them and they enter into an electrovalent bond.

Hence;

Be²⁺ + 2F⁻ → BeF₂

8 0

2 years ago