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Brrunno [24]
1 year ago
14

Calculate the mass of 3.4 moles of nitric acid (hno3). explain the process or show your work by including all values used to det

ermine the answer.
Chemistry
1 answer:
balu736 [363]1 year ago
6 0

Answer:

6.34917360^25g

Explanation:

It's been a while since I've done this type of problem so I'm not making any promises that its right hahaha, but I hope it helps anyway. Please let me know whether I'm right or not!

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A gas that occupies 50.0 liters has its volume increased to 68 liters when the pressure was changed to 3.0 ATM. What was the ori
Yuliya22 [10]

Answer:

4.1 atm = 3,116 mmHg = 415.4 kPa

Explanation:

According to Boyle's law, as volume is increased the pressure of the gas is decreased. That can be expressed as:

P₁ x V₁= P₂ x V₂

Where P₁ and V₁ are the initial pressure and volume respectively, and P₂ and V₂ are final pressure and volume, respectively.

From the problem, we have:

V₁= 50.0 L

V₂= 68.0 L

P₂= 3.0 atm

Thus, we calculate the initial pressure as follows:

P₁= (P₂ x V₂)/V₁= (3.0 atm x 68.0 L)/(50.0 L)= 4.08 atm ≅ 4.1 atm

To transform to mmHg, we know that 1 atm= 760 mmHg:

4.1 atm x 760 mmHg/1 atm = 3,116 mmHg

To transform to kPa we use: 1 atm= 101.325 kPa

4.1 atm x 101.325 kPa = 415.4 kPa

5 0
2 years ago
Which form of emission is commonly not written in nuclear equations because they do not affect charges, atomic numbers, or mass
oksano4ka [1.4K]

Answer: It is Gamma ray

4 0
3 years ago
Read 2 more answers
Which type of energy transformation comes from electric heater?​
ivolga24 [154]

Answer:

electric energy ---> heat energy or A.

Explanation:

the name says it all

3 0
3 years ago
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A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solution
Dmitrij [34]

Answer:

The change in internal energy is - 1.19 kJ

Explanation:

<u>Step 1:</u> Data given

Heat released = 3.5 kJ

Volume calorimeter = 0.200 L

Heat release results in a 7.32 °C

Temperature rise for the next experiment = 2.49 °C

<u>Step 2:</u> Calculate Ccalorimeter

Qcal = ccal * ΔT ⇒ 3.50 kJ = Ccal *7.32 °C

Ccal = 3.50 kJ /7.32 °C = 0.478 kJ/°C

<u>Step 3:</u> Calculate energy released

Qcal = 0.478 kJ/°C *2.49 °C = 1.19 kJ

<u>Step 4:</u> Calculate change in internal energy

ΔU =  Q + W       W = 0  (no expansion)

Qreac = -Qcal = - 1.19 kJ

ΔU = - 1.19 kJ

The change in internal energy is - 1.19 kJ

4 0
3 years ago
Which property of a substance is not altered by a physical change?
antiseptic1488 [7]
C. Composition

is the answer
3 0
3 years ago
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