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3 years ago

HURRY WILL MARK BRAINILEST

Chemistry

2 answers:

Mnenie [13.5K]3 years ago

6 0

Answer:

Give Brainliest Please!

natita [175]3 years ago

5 0

Answer:

Explanation:

diffusion has already occurred

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All of the following are reasons why equations or balanced EXCEPT

Anni [7]

Answer:

monkey

Explanation:

4 0

2 years ago

How many moles of gas would occupy 22.4 L at 273K in one arm?

andrew-mc [135]

Answer:

1 mol

Explanation:

Using the general gas law equation as follows:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

According to the provided information in the question;

V = 22.4L

T = 273K

P = 1 atm

R = 0.0821 Latm/molK

n = ?

Using PV = nRT

n = PV/RT

n = (1 × 22.4) ÷ (0.0821 × 273)

n = 22.4 ÷ 22.4

n = 1mol

4 0

2 years ago

B) How many kilograms of carbon dioxide are formed when 24.42 g of iron is<br> produced?

charle [14.2K]

Answer:

0.0289 kg of CO2 will be formed

Explanation:

Step 1: Data given

Mass of iron produced = 24.42 grams

Atomic mass iron = 55.845 g/mol

Molar mass CO2 = 44.01 g/mol

Step 2: The balanced equation

Fe2O3 + 3CO → 2Fe + 3CO2

Step 3: Calculate moles iron

Moles iron = mass iron / molar mass iron

Moles iron = 24.42 grams / 55.845 g/mol

Moles iron = 0.437 moles

Step 4: Calculate moles CO2

For 1 mol Fe2O3 we need 3 moles CO to produce 2 moles Fe and 3 moles CO2

For 0.437 moles Fe we'll have 3/2 * 0.437 = 0.6555 moles CO2

Step 5: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.6555 moles * 44.01 g/mol

Mass CO2 = 28.85 grams = 0.0289 kg

0.0289 kg of CO2 will be formed

5 0

3 years ago

How many grams of agcl would be needed to make a 4. 0 m solution with a volume of 0. 75 l?

devlian [24]

430 g of AgCl would be needed to make a 4.0m solution with a volume of 0.75 L.

<h3>What is Molarity?</h3>

The amount of a substance in a specific volume of solution is known as its molarity (M).

The number of moles of a solute per liter of a solution is known as molarity.

<h3>Calculation of Required amount of AgCl</h3>

Remember that mol/L is the unit of molarity (M).

We can compute the necessary number of moles of solute by multiplying the concentration by the liters of solution, according to dimensional analysis.

0.75L×4.0M=3.0mol

Then, using the periodic table's molar mass for AgCl, convert from moles to grams:

3.0mol×143.321gmol=429.963g

The final step is to round to the correct significant figure, which in this case is two: 430g.

Hence, 430 g of AgCl would be needed to make a 4.0m solution with a volume of 0.75 L.

Learn more about Molarity here:
brainly.com/question/8732513

#SPJ4

8 0

1 year ago

Iron(III) oxide and hydrogen react to form iron and water, like this: Fe_2O_3(s) + 3H_2(g) rightarrow 2Fe(s) + 3H_2O(g) At a cer

Sergeeva-Olga [200]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The equilibrium constant is $K_c= 2.8*10^{-4}$

Explanation:

From the question we are told that

The chemical reaction equation is

$Fe_{2} O_{3}_{(s)} + 3H_{2}_{(g)} -----> 2Fe_{(s)} + 3H_{2} O_{(g)}$

The voume of the misture is $V_m = 5.4L$

The molar mass of $Fe_{2} O_{3}_{(s)}$ is a constant with value of $M_{Fe_{2} O_{3}_{(s)} } = 160g/mol$

The molar mass of $H_{2}_{(g)}$ is a constant with value of $H_2 = 2g/mol$

The molar mass of $H_{2}O$ is a constant with value of $H_2O = 18g/mol$

Generally the number of moles is mathematically given as

$No \ of \ moles \ = \frac{mass}{molar\ mass}$

For $Fe_{2} O_{3}_{(s)}$

$No \ of\ moles = \frac{3.54}{160}$

$= 0.022125 \ mols$

For $H_{2}$

$No \ of\ moles = \frac{3.63}{2}$

$= 1.815 \ mols$

For $H_{2}O$

$No \ of\ moles = \frac{2.13}{18}$

$= 0.12 \ mols$

Generally the concentration of a compound is mathematicallyrepresented as

$Concentration = \frac{No \ of \ moles }{Volume }$

For $Fe_{2} O_{3}_{(s)}$

$Concentration[Fe_2 O_3] = \frac{0.222125}{5.4}$

$= 4.10*10^{-3}M$

For $H_{2}$

$Concentration[H_2] = \frac{1.815}{5.4}$

$= 0.336M$

For $H_{2}O$

$Concentration [H_2O] = \frac{0.12}{5.4}$

$= 0.022M$

The equilibrium constant is mathematically represented as

$K_c = \frac{[concentration \ of \ product]}{[concentration \ of \ reactant ]}$

Considering $H_2O \ for \ product$

And $H_2 \ for \ reactant$

At equilibrium the

$K_c = \frac{0.022}{0.336}$

$K_c= 2.8*10^{-4}$

3 0

3 years ago