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FromTheMoon [43]
3 years ago
11

Mass of a substance divided by unit volume is called

Chemistry
1 answer:
olga2289 [7]3 years ago
5 0
Answer is its density
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What is the density of an object with a mass of 1.663 g and a volume of 0.2009 mL
daser333 [38]
<span>8.278 g/mL The definition of density is mass per volume. So what you need to do is divide the known mass by the known volume. So 1.663 g / 0.2009 mL = 8.27775 g/mL But you also have to keep track of significant figures. Since both 1.663 and 0.2009 have 4 significant figures each, you need to round the result to 4 significant figures. So 8.27775 g/mL = 8.278 g/mL</span>
7 0
3 years ago
Read 2 more answers
I need help with 1,2,3, and 4
Schach [20]

Answer:

  • Problem 1: 1.85atm
  • Problem 2: 110mL
  • Problem 3: 290 mL
  • Problem 4: 1.14 atm

Explanation:

Problem 1

<u>1. Data</u>

<u />

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

<u>2. Formula</u>

Since the temeperature is constant you can use Boyle's law for idial gases:

          PV=constant\\\\P_1V_1=P_2V_2

<u>3. Solution</u>

Solve, substitute and compute:

         P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2

        P_2=3.25atm\times755mL/1325mL=1.85atm

Problem 2

<u>1. Data</u>

<u />

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

<u>2. Formula</u>

You assume that the temperature does not change, and then can use Boyl'es law again.

          P_1V_1=P_2V_2

<u>3. Solution</u>

This time, solve for V₂:

           P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2

Substitute and compute:

        V_2=548mmHg\times 125mL/625mmHg=109.6mL

You must round to 3 significant figures:

        V_2=110mL

Problem 3

<u>1. Data</u>

<u />

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

<u>2. Formula</u>

At constant pressure, Charle's law states that volume and temperature are inversely related:

         V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

The temperatures must be in absolute scale.

<u />

<u>3. Solution</u>

a) Convert the temperatures to kelvins:

  • T₁ = 25 + 273.15K = 298.15K

  • T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

        \dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml

You must round to two significant figures: 290 ml

Problem 4

<u>1. Data</u>

<u />

a) P = 865mmHg

b) Convert to atm

<u>2. Formula</u>

You must use a conversion factor.

  • 1 atm = 760 mmHg

Divide both sides by 760 mmHg

       \dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}

<u />

<u>3. Solution</u>

Multiply 865 mmHg by the conversion factor:

    865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer

3 0
3 years ago
A student is doing online research. The student's teacher said that some
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Websites that end in .com or .org should be avoided.
3 0
3 years ago
Describe the relationship between the light-dependent and the light-independent reactions.
Slav-nsk [51]
<span>The products of the light-dependent reactions are used to help 'fuel' the light-independent reactions. 

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NADPH and ATP are produced during the light-dependent reaction for use in the light-independent reaction (the Calvin Cycle). </span>
3 0
3 years ago
Mr. Ragusa asks Hassan to make silver crystals from the following reaction.
SVEN [57.7K]

Answer:

Percentage yield = 61.7%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated:

2AgNO₃ + Cu —> Cu(NO₃)₂ + 2Ag

Next, we shall determine the mass of AgNO₃ that reacted and the mass of Ag produced from the balanced equation. This is illustrated below:

Molar mass of AgNO₃ = 108 + 14 + (16×3)

= 108 + 14 + 48

= 170 g/mol

Mass of AgNO₃ from the balanced equation = 2 × 170 = 340 g

Molar mass of Ag = 108 g/mol

Mass of Ag from the balanced equation = 2 × 108 = 216 g

SUMMARY:

From the balanced equation above,

340 g of AgNO₃ reacted to produce 216 g of Ag.

Next, we shall determine the theoretical yield of Ag. This can be obtained as follow:

From the balanced equation above,

340 g of AgNO₃ reacted to produce 216 g of Ag.

Therefore, 51 g of AgNO₃ will react to produce = (51 × 216)/340 = 32.4 g of Ag.

Thus, the theoretical yield of Ag is 32.4 g.

Finally, we shall determine the percentage yield of Ag. This can be obtained as follow:

Actual yield = 20 g

Theoretical yield = 32.4 g

Percentage yield =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield = 20 / 32.4 × 100

Percentage yield = 61.7%

6 0
3 years ago
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