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ch4aika [34]
2 years ago
4

What is the volume in liters of 321 g of a liquid with a density of 0.84 g/mL?

Chemistry
1 answer:
mars1129 [50]2 years ago
8 0

Answer:

  • <u>0.38 liter</u>

Explanation:

<u>1) Data:</u>

a) V = ?

b) m = 321 g

c) d = 0.84 g/ml

<u>2) Principles and formulae:</u>

Density measures the ratio of the mass and the volume of the substances:

  • d = m / V

Where:

  • d is density of the substance
  • m is the mass of the substance
  • V is the volume occupied by the substance

<u>3) Calculations:</u>

Solve for V, substitute the data in the formua, and compute:

  • d = m / V ⇒ V = m / d
  • V = 321 g / 0.84 g/ml = 382.14 ml

Convert ml to liter:

  • 382.14 ml × 1 liter / 1,000 ml = 0.38 liter (rounded to two decimal places).
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A 20.0 -\,L volume of an ideal gas in a cylinder with a piston is at a pressure of 3.2 atm. Enough weight is suddenly removed fr
zzz [600]

Answer:

1. ΔE = 0 J

2. ΔH = 0 J

3. q = 3.2 × 10³ J

4. w = -3.2 × 10³ J

Explanation:

The change in the internal energy (ΔE) and the change in the enthalpy (ΔH) are functions of the temperature. If the temperature is constant, ΔE = 0 and ΔH = 0.

The gas initially occupies a volume V₁ = 20.0 L at P₁ = 3.2 atm. When the pressure changes to P₂ = 1.6 atm, we can find the volume V₂ using Boyle's law.

P₁ × V₁ = P₂ × V₂

3.2 atm × 20.0 L = 1.6 atm × V₂

V₂ = 40 L

The work (w) can be calculated using the following expression.

w = - P . ΔV

where,

P is the external pressure for which the process happened

ΔV is the change in the volume

w = -1.6 atm × (40L - 20.0L) = -32 atm.L × (101.325 J/1atm.L) = -3.2 × 10³ J

The change in the internal energy is:

ΔE = q + w

0 = q + w

q = - w = 3.2 × 10³ J

6 0
2 years ago
A student has 100. mL of 0.400 M CuSO4 (aq) and is asked to make 100. mL of 0.150 M CuSO4 (aq) for a spectrophotometry experimen
GenaCL600 [577]

Answer:

We have to take 37.5 mL of a 0.400 M solution

Explanation:

Step 1: Data given

Stock volume = 100 mL  = 0.100L

Stock concentration 0.400 M

Volume of solution he wants to make = 100 mL = 0.100L

Concentration of solution he wants to make = 0.150 M

Step 2: Calculate the volume of 0.400 M CuSO4 needed

C1*V1 = C2*V2

⇒with C1 = the stock concentration = 0.400M

⇒with V1 = the volume of the stock = TO BE DETERMINED

⇒with C2 = the concentration of the solution he wants to make = 0.150 M

⇒with V2 = the volume of the solution made = 0.100 L

0.400 M * V1 = 0.150M * 0.100L

V1 = (0.150M*0.100L) / 0.400 M

V1 = 0.0375 L = 37.5 mL

We have to take 37.5 mL of a 0.400 M solution

5 0
2 years ago
Read 2 more answers
Consider the following generic chemical equation:
vredina [299]

Answer:  238.6 J

Explanation:

According to the law of conservation of energy, energy can neither be created nor be destroyed. It can only be transformed from one form to another.

Endothermic reactions are those in which heat is absorbed by the system and thus the energy of products is higher than the energy of reactants.

For the given reaction:

A+B+energy\rightarrow C+D

Energy of A = 85.1 J

Energy of B = 87.9 J

Energy on reactant side =  Energy of A + Energy of B  + Energy absorbed 85.1 + 87.9 + 104.3 = 277.3 J

Energy on reactant side = Energy on product side = 277.3 J

Energy on product side = Energy of C + Energy of D

277.3 J = 38.7 J  + Energy of D

Energy of D = 238.6 J

Thus chemical energy product D must contain is 238.6 J

6 0
2 years ago
Read 2 more answers
Milk, if 2.00 liters has a mass of 2.06 kg. Find the density in g/cm^3
Gre4nikov [31]

Answer:

<h3>The answer is 1.03 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

1 kg = 1000 g

2.06 kg = 2060 g

1 L = 1000 cm³

2 L = 2000 cm³

mass = 2060 g

volume = 2000 cm³

We have

density =  \frac{2060}{2000}  =  \frac{103}{100}  \\

We have the final answer as

<h3>1.03 g/cm³</h3>

Hope this helps you

5 0
2 years ago
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A 62.6-gram piece of heated limestone is placed into 75.0 grams of water at 23.1°C. The limestone and the water come to a final
alekssr [168]

Answer:

208.7°C was the initial temperature of the limestone.

Explanation:

Heat lost by limestone will be equal to heat gained by the water

-Q_1=Q_2

Mass of limestone = m_1=62.6 g

Specific heat capacity of limestone = c_1=0.921 J/g^oC

Initial temperature of the limestone = T_1=?

Final temperature = T_2=T =  51.9°C

Q_1=m_1c_1\times (T-T_1)

Mass of water= m_2=75.0 g

Specific heat capacity of water= c_2=4.186 J/g^oC

Initial temperature of the water = T_3=23.1^oC

Final temperature of water = T_2=T =  51.9°C

Q_2=m_2c_2\times (T-T_3)

-Q_1=Q_2

-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)

On substituting all values:

-(62.6 g\times 0.921 J/g^oC\times (51.9^oC-T_1))=75.0 g\times 4.186 J/g^oC\times (51.9^oC-23.1^oC)

T_1=208.7^oC

208.7°C was the initial temperature of the limestone.

7 0
2 years ago
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