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2 years ago

ASAP

Chemistry

1 answer:

sammy [17]2 years ago

8 0

The answer is C.

The vast difference in electronegativity of the oxygen and hydrogen in water, the O-H bond is polar.

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4.3 moles of a gas are at a temperature of 28 degrees * C with a pressure of 1.631 atm. What volume does the gas occupy?

Shkiper50 [21]

Answer:

65.2L

Explanation:

Using the general gas equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (Litres)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (Kelvin)

According to the information provided in this question,

P = 1.631 atm

V = ?

n = 4.3 moles

T = 28°C = 28 + 273 = 301K

Using PV = nRT

V = nRT/P

V = 4.3 × 0.0821 × 301 ÷ 1.631

V = 106.26 ÷ 1.631

V = 65.15

Volume of the gas = 65.2L

7 0

3 years ago

Which of the following is a gaseous substance that is liquid or solid at room

Sveta_85 [38]

Answer:

liquid is the correct answer

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2 years ago

Please help. im freaking out rn. i have like 40 missing assignments please

Katarina [22]

Answer:

I'm pretty sure its the one that says very little at the beginning but if I get it wrong I'm sorry

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3 years ago

Which organisms are the best example of producers in a food web? A. zooplankton B. krill C. killer whales D. marine phytoplankto

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D. Marine phytoplankton

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3 years ago

Read 2 more answers

The chemical equation shows iron(III) phosphate reacting with sodium sulfate. 2FePO4 + 3Na2SO4 Fe2(SO4)3 + 2Na3PO4 What is the t

slava [35]

<u>Answer:</u> The theoretical yield of iron(III) sulfate is 26.6 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)

Given mass of iron(III) phosphate = 20.00 g

Molar mass of iron(III) phosphate = 150.82 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron(III) phosphate}=\frac{20g}{150.82g/mol}=0.133mol$

The given chemical equation follows:

$2FePO_4+3Na_2SO_4\rightarrow Fe_2(SO_4)_3+2Na_3PO_4$

As, sodium sulfate is present in excess. So, it is considered as an excess reagent.

Thus, iron(III) phosphate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of iron(III) phosphate produces 1 mole of iron(III) sulfate

So, 0.133 moles of iron(III) phosphate will produce = $\frac{1}{2}\times 0.133=0.0665moles$ of iron(III) sulfate

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

Molar mass of iron(III) sulfate = 399.9 g/mol

Moles of iron(III) sulfate = 0.0665 moles

Putting values in equation 1, we get:

$0.0665mol=\frac{\text{Mass of iron(III) sulfate}}{399.9g/mol}\\\\\text{Mass of iron(III) sulfate}=(0.0665mol\times 399.9g/mol)=26.6g$

Hence, the theoretical yield of iron(III) sulfate is 26.6 grams

8 0

3 years ago