What is the molarity of a 300.0 mL solution containing 25.0 g of NaCl?

The stage of engine operation when both the intake and exhaust valves are closed is the _______ stage. A. compression B

Varvara68 [4.7K]

D. the exhaust valve

3 0

3 years ago

What is the amount of aluminum chloride produced from 40 moles of aluminum and excess chlorine?

Marrrta [24]

Answer:

is it 26.98

Explanation:

6 0

2 years ago

What amount of energy is required to change a spherical drop of water with a diameter of 1.80 mm to three smaller spherical drop

Gekata [30.6K]

This is a straightforward question related to the surface energy of the droplet.

You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. 

With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. 

The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ 

The five smaller droplets need to have the same volume as the original. Therefore 

5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. 

Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. 

Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². 

The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ 
From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. 

Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets.

7 0

3 years ago

Which of the following would be considered important in including in the summary of an experimental investigation?

just olya [345]

Answer is: variables that were manipulated during the experiment.

The two main variables in an experiment are the independent and dependent variable.

Dependent variable is the variable being tested and measured in a scientific experiment.

Dependent variables depend on the values of independent variables. The dependent variables represent the output or outcome whose variation is being studied.

Laboratory room number is the least important.

7 0

3 years ago

Read 2 more answers

A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35 oC to 195 oC. Calculate the specific heat.

Alja [10]

Answer:

Question 1: 0.2J/(gºC)

Question 2: 6,000J

Question 3: 300J

Question 4: 80g

Question 5: 74ºC

Question 6: 50g

Explanation:

Question 1. A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35º oC to 195º C. Calculate the specific heat.

The thermal energy equation is:

Q = m × C × ΔT

Substitute and solve for C:

566J = 20g × C × (195ºC - 35ºC)

C = 566J / (20g × 160ºC)

C = 0.177 J/(gºC) ≈ 0.2J/(gºC)

You must round to one significant figure because one factor has one significant figure).

Qustion 2. 40g of water is heat at 40ºC and the temperature rise to 75ºC. What is the amount of heat needed for the temperature to rise? (specific heat of water is 4.184 J/gºC)

Use the thermal energy equation again:

Q = m × C × ΔT

Substitute and compute:

Q = 40g × 4.184 J/gºC × (75ºC - 40ºC)

Q = 5,857.6J

Round to one significant figure: 6,000J

Question 3. Graphite has a mass of 50g and a specific heat of 0.420 J/gºC. If graphite is cooled from 50ºC to 35ºC, how much energy was lost?

Q = m × C × ΔT

Q = 50g × 0.420J/gºC × (35ºC - 50ºC)

Q = 315J

Round to one significant figure (because 50g has one significant figure)

Q = 300J

Question 4. Iron has a specific heat of 0.712 J/gºC. A piece of iron absorbs 3000J of energy and undergoes a temperature change totaling 50ºC, What is the mass of iron?

Q = m × C × ΔT

Solve for m:

m = Q / (C × ΔT)

Substitute and compute:

m = 3,000J / (0.712J/gºC × 50ºC)

m = 84.26 g ≈ 80 g (rounded to one significant figure, because the factor 3,000J has one significant figure).

Question 5. If 400g of an unknown solution at 70ºC loses 7500 J of heat, what is the final temperature of the unknown solution. The unknown solution has a specific heat of 4.184 J/gºC.

Q = m × C × ΔT

Q is negative, since it is released.

Substitute and solve for T:

- 7,500J = 400g × 4.184J/gºC × (T - 70ºC)

T = - 7500J / 400g × 4.184J/gºC) + 70ºC

T = 74ºC

If you round to one significant figure you cannot tell the temperature difference, thus leave two significant figures.

Question 6. How many grams of water would require 9500J of heat to raise the temperature from 50ºC to 100ºC

Q = m × C × ΔT

Subsitute:

9,500J = m × 4.184J/gºC × (100ºC - 50ºC)

Solve for m and compute:

m = 9,500J / (4.184J/gºC × 50ºC)

m = 45g

Since the temperatures indicate one singificant figure, the mass should be rounded to one significant figure:

m = 50g.

8 0

2 years ago