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mr_godi [17]
2 years ago
14

If you use 19 mL of 0.100 M KMnO4, how many moles have you used? The answer has to be in four decimal places (do not forget the

leading zero). No scientific notation.
How much 5.0 M H2SO4 needs to be used to make 152 mL of 1.0 M? Round answer to nearest whole number and omit units.
Chemistry
1 answer:
Scorpion4ik [409]2 years ago
3 0
1)

Volume = 19 / 1000 = 0.019 L

n = M * V

n = 0.100 * 0.019

n = 0.0019 moles
__________________________

2)

 M₁ * V₁ = M₂* V₂

5.0 * V₁ = 1.0 * 152

5.0 V₁ = 152

V₁ = 152 / 5.0

V₁ = 30.4 mL
______________________________________

<span>hope this helps!</span>
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What are all the intermolecular forces present in a sample of BrF?
yuradex [85]

Answer:

dipole-dipole

Explanation:

Intermolecular forces exists between the molecules of a substance in a particular state of matter.

The type of intermolecular forces present in  a substance is determined by the electronegativity difference between the atoms that compose the substance.

There is a non zero electronegativity difference  between Br and F hence the molecule is polar and the intermolecular forces between the molecules of BrF are dipole-dipole forces.

4 0
2 years ago
A force of 3000N is applied on a 20kg ball. Find the acceleration of the ball
Andrews [41]

Answer:

force = 3000N

mass= 20 kg

now

F= ma

3000= 20×a

3000÷20=a

a=15

F= ma by newtons second law of motion

3 0
2 years ago
This element exists in adundance in the sun.Explain how you would go about capturing sunlight.Would this captured sunlight conta
nasty-shy [4]
Yes because the sun will make alots of thing grow
4 0
3 years ago
What is the pressure of 0.5 mol of nitrogen gas in a 5 L container at 203 K
vlada-n [284]

Answer:

=1.666 liters

Explanation:

1 mole of a has at standard temperature and pressure occupies a volume of 22.4 liters.

0.5 moles of nitrogen occupy a volume of (0.5 moles×22.4 dm³/mol)/ 1

=11.2 liters.

Standard pressure= 1 atmosphere (Atm)

Standard temperature = 273.15 Kelvin

According to Combined gas equation, P₁V₁/T₁=P₂V₂/T₂

Let us take the conditions under standard conditions as the reference, with the subscript 1 and the conditions under the 5L container to be scenario 2 with subscript 2.

Therefore P₂ =P₁V₁T₂/T₁V₂

Substituting for the values we get:

P₂= (1 atm× 11.2L ×203K)/ (273K×5L)

=1.666 atm

5 0
3 years ago
If you have a 1500 g aluminum pot, how much heat energy is needed to raise its temperature by 100°C?
Nataly [62]

The heat energy required to raise the temperature of 1500 g of aluminium pot by 100°C is 135 kJ.

The heat energy required to raise the temperature of 1500 g of copper pot by 100 °C is 57.75 kJ.

Explanation:

The heat energy required to raise the temperature of any body can be obtained from the specific heat formula. As this formula states that the heat energy required to raise the temperature of the body is directly proportional to the product of mass of the body, specific heat capacity of the material and temperature change experienced by the material.

So in this problem, the mass of the aluminium is given as m = 1500 g, the specific heat of the aluminium is 0.900 J/g °C. Then as it is stated that the temperature is raised by 100 °C, so the pots are heat to increase by 100 °C from its initial temperature. This means the difference in temperature will be 100°C (ΔT = 100°C).

Then, the heat energy required to raise the temperature will be

q = m*c*del T = 1500 * 0.900 * 100 = 135000 = 135 kJ

Thus, the heat energy required to raise the temperature of 1500 g of aluminium pot by 100 °C is 135 kJ.

Similarly, the mass of copper pot is given as 1500 g, the specific heat capacity of copper is 0.385 and the difference in temperature is 100  °C.

Then, the heat energy required to raise its temperature will be

q = m*c*del T = 1500 * 0.385 * 100 = 57750 = 57.75 kJ

And the heat energy required to raise the temperature of 1500 g of copper pot by 100°C is 57.75 kJ.

So, the heat energy required to raise the temperature of 1500 g of aluminium pot by 100°C is 135 kJ. And the heat energy required to raise the temperature of 1500 g of copper pot by 100 °C is 57.75 kJ.

8 0
3 years ago
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