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Anestetic [448]
3 years ago
10

The energy of motion is known as potential energy. Question 1 options: True False

Chemistry
1 answer:
Korolek [52]3 years ago
6 0

Answer:

False

Explanation:

This is because potential energy is energy that it not used yet.

You might be interested in
what is the correct conversion factor by which to multiply to convert moles of nitrogen to moles of ammonium nitrate (NH4NO3)
Ostrovityanka [42]

Answer:

0.5

Explanation:

1 mole of ammonium nitrate contains 2 moles of nirogen

1 mole of nitrogen converts to 0.5  moles of ammonium nitrate

the conversation factor is 0.5

6 0
3 years ago
Read 2 more answers
Which of the following is an ion with a correct charge ?<br><br> O 4-<br> Ne8+<br> Ca2+<br> Na1-
KengaRu [80]

Answer:

Ca2+ is an ion with a correct charge.

8 0
3 years ago
a sample of a gas has a volume kf 640 cm^{3} at 100°c and 1490 mmhg,what would be its volume at stp?​
scoundrel [369]

Answer:

V₂ = 918.1 cm³

Explanation:

Given data:

Initial volume = 640 cm³

Initial temperature = 100°C (100+273 = 373 K)

Initial pressure = 1490 mmHg (1490 /760 = 1.96 atm)

Final volume = ?

Final temperature = 273 K

Final pressure = 1 atm

Solution:

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

now we will put the values in formula.

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 1.96 atm × 640 cm³ × 273 K / 373 K × 1 atm

V₂ = 342451.2 atm .cm³ . K / 373 K. atm

V₂ = 918.1 cm³

3 0
3 years ago
2 Nobr +heat &gt; 2 no2+br2 what happens when you remove nobr
Kryger [21]
There would be no nobr and it would just be 2
6 0
3 years ago
250 ml of seawater and we inked each molecule with pink color, then we mixed this 250 ml in the ocean. After mixing you took 250
Marysya12 [62]

Answer:

9.77 × 10⁹ molecules

Explanation:

Since the density of water is 1 g/cm³ = 1000 g/L

So, we find the mass of sea water in 250 mL = 0.250 L

We know density = mass/volume

mass = density × volume = 1000 g/L × 0.250 L = 250 g

Now we calculate the number of moles of sea water in 250 g, 250 mL of sea water.

number of moles n = mass of sea water,m/molar mass of water, M

molar mass of water, M = 18 g/moL

n = m/M = 250 g/18 g/mol = 13.89 mol

We now calculate the number of molecules present in the 250 mL of sea water.

n = N/N' where N = number of molecules, N' = avogadro's number = 6.022 × 10²³/mol

So,N =nN' = 13.89 mol × 6.022 × 10²³/mol = 8.36 × 10²⁴ molecules

Now, the volume of the ocean is 1.337 × 10¹⁸ m³ = 1.337 × 10¹⁵ L

Since the 250 mL sea water is mixed with the ocean, the number of molecules per liter is 8.36 × 10²⁴ molecules/1.337 × 10¹⁵ L = 6.25 × 10⁹ molecules/L

If we now take 250 mL out of the ocean, the number of molecules in this 250 mL will be 6.25 × 10⁹ molecules/L × 250 mL =  6.25 × 10⁹ molecules/L × 0.250 L = 9.77 × 10⁹ molecules.

So, we have 9.77 × 10⁹ molecules of pink molecules in the 250 mL after mixing in the ocean.

6 0
3 years ago
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