CHEMISTRY HIGH SCHOOL

Which of the following statements is not correct?0.20 mole of O2 = 6.4 g
0.75 mole of H2CO3 = 47 g
3.42 moles CO = 95.8 g
4.1 moles Li2O = 94 g

Answers

Answer 1

Answer:

4.1 moles Li2O = 94 g (incorrect)

4.1 mole of Li2O = 122.6g(correct)

Explanation:

1. Molar mass of O2 = 32

0.2 mole of 32 = 0.2×32 = 6.4g (correct)

2. Molar mass of H2CO3 = 62.03

0.75 mole of 62.03 = 0.75×62.03 = 46.5g (47 is correct)

3. Molar mass of CO = 28

3.42 moles of 30 = 3.42×28 = 95.76 (95.8 is correct)

4. Molar mass of Li2O = 29.9

4.1 mole of Li2O = 4.1×29.9 = 122.59g (94g is incorrect)

Answer 2

Answer:

Answer:

It's 4.1 moles Li2O = 94 g on Odyssey ware

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chlorine has 7 outer energy level electrons and argon has 8. what is the difference in the reactivity of these two elements?

Answers

chlorine is more reactive than argon, because argon is not reactive at all. Chlorine only has to gain 1 valence electron

Chlorine is highly reactive as it seeks to gain an extra electron to balance its outer shell (max 8) and argon is stable because it has a stable outer shell

A nitride ion has 7 protons, 8 neutrons, and 10 electrons. What is the overall charge on this ion?

Answers

The overall total would be (3-) charge. Hope I helped./

The overall total would be (3-) charge

Which of these pairs of elements have the same number of valence electrons?hydrogen (H) and helium (He)
sodium (Na) and potassium (K)
carbon (C) and oxygen (O)
boron (B) and beryllium (Be)

Answers

The correct answer is the second option; sodium (Na) and potassium (K.)
Both sodium and potassium have the same number of valence electrons.

Answer:

The correct answer is the second option; sodium (Na) and potassium (K.)

Both sodium and potassium have the same number of valence electrons.

Explanation:

Calculate the volume of each of the following gases at STP 7.6 mol Ar 0.44 mol C2,H3

Answers

For Ar :

1 mol ------------ 22.4 L ( at STP )
7.6 mol ---------- x L

x = 7.6 * 22.4

x = 170.24 L
-----------------------------------------------------------------
For C2H3:

1 mol ------------ 22.4 ( at STP)
0.44 mol --------- y L

y = 0.44 * 22.4

y = 9.856 L

hope this helps !.

The volume of 0.44 moles of ethylene gas at STP is approximately 10.33 liters.

To calculate the volume of gases at Standard Temperature and Pressure (STP), we can use the ideal gas law:

PV = nRT

Where:

P = Pressure (at STP, it's 1 atmosphere, or 1 atm)

V = Volume (in liters)

n = Number of moles

R = Ideal gas constant (0.0821 L·atm/mol·K)

T = Temperature (at STP, it's 273.15 K)

Let's calculate the volume for each gas:

For 7.6 moles of Ar (argon):

P = 1 atm

n = 7.6 moles

R = 0.0821 L·atm/mol·K

T = 273.15 K

Now, plug these values into the ideal gas law:

V = (nRT) / P

V = (7.6 moles * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm

V ≈ 172.75 liters

So, the volume of 7.6 moles of argon gas at STP is approximately 172.75 liters.

For 0.44 moles of C2H3 (ethylene):

P = 1 atm

n = 0.44 moles

R = 0.0821 L·atm/mol·K

T = 273.15 K

Now, plug these values into the ideal gas law:

V = (nRT) / P

V = (0.44 moles * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm

V ≈ 10.33 liters

For more such question on ethylene gas visit:

brainly.com/question/3929010

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Answers

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