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Anna35 [415]
3 years ago
6

Imagine you have an object with a non-geometric shape having a mass of 26.7 g and

Chemistry
1 answer:
Bas_tet [7]3 years ago
7 0

Answer:

The new reading of the water level is 54.81 mL.

Explanation:

From the question given above, the following data were obtained:

Mass of object = 26.7 g

Density of object = 0.767 g/mL

Volume of water = 20 mL

New reading of the water level =?

Next, we shall determine the volume of the object. This can be obtained as follow:

Mass of object = 26.7 g

Density of object = 0.767 g/mL

Volume of object =?

Density = mass /volume

0.767 = 26.7 / volume of object

Cross multiply

0.767 × Volume of object = 26.7

Divide both side by 0.767

Volume of object = 26.7 / 0.767

Volume of object = 34.81 mL

Finally, we shall determine the new reading of the water level. This can be obtained as follow:

Volume of water = 20 mL

Volume of object = 34.81 mL

New reading of the water level =?

New reading of the water level = (Volume of water) + (Volume of object)

New reading = 20 + 34.81

New reading of the water level = 54.81 mL

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8 0
4 years ago
Your general impression suggests that your immediate action should be to: ____.
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Your general impression suggests that your immediate action should be to call for additional help. The correct option is a.

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During the time of general impression, one should immediately call for additional help. Additional people will give you suggestions for the problem.

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6 0
1 year ago
Phosphorous trichloride and phosphorous pentachloride equilibrate in the presence of molecular chlorine according to the reactio
umka21 [38]

Answer : The value of K_p at this temperature is 66.7

Explanation : Given,

Pressure of PCl_3 at equilibrium = 0.348 atm

Pressure of Cl_2 at equilibrium = 0.441 atm

Pressure of PCl_5 at equilibrium = 10.24 atm

The balanced equilibrium reaction is,

PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{PCl_5})}{(p_{PCl_3})(p_{Cl_2})}

Now put all the values in this expression, we get :

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K_p=66.7

Therefore, the value of K_p at this temperature is 66.7

4 0
3 years ago
How much energy is required to heat 36.0 g H2O from a liquid at 65°C to a gas at 115°C? The following physical data may be usefu
In-s [12.5K]

Answer:

energy required=qnet=87.75kJ

Explanation:

we will do it in three seperate step and then add up those value.

first step is to heat the sample of water upto 100C i.e upto boiling pont. because just after this sample of water started vaporization.

q 1= m c (T2-T1)

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next is to vaporize the sample at 100C

q2 = 36.0 g / 18.0 g/mol X 40.7 kJ/mol

q2= 81.4 kJ

Finally, heat the steam upto 115C

q3 = m c (T2-T1)

q 3= 36.0 g (2.01 J/gC)(115-100C)

q3 = 1085 J =1.085kJ

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Answer:

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