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3 years ago

Do viruses have cell walls?

Chemistry

1 answer:

stiks02 [169]3 years ago

4 0

No, they have a protein coat that keeps the virus in.

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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

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

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

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

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

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

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

$P_1V_1=P_2V_2$

3. Solution

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

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

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.

3. Solution

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

1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

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

3. Solution

Multiply 865 mmHg by the conversion factor:

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

3 0

3 years ago

A sample of hydrated copper (II) sulfate (CuSO4•nH2O) is heated to 150°C and produces 103.74 g anhydrous copper (II) sulfate and

wariber [46]

Answer:

Explanation:

Firstly, we convert what we have to percentage compositions.

There are two parts in the molecule, the sulphate part and the water part.

The percentage compositions is as follows:

Sulphate- (103.74)/(103.74 + 58.55) × 100% = apprx 64%

The water part = 100 - 64 = 36%

Now, we divide the percentages by the molar masses.

For the CuSO4 molar mass is 64 + 32 + 4(16) = 160g/mol

For the H2O = 2(1) + 16 = 18g/mol

Now we divide the percentages by these masses

Sulphate = 64/160 = 0.4

Water = 36/18 = 2

The ratio is thus 0.4:2 = 1:5

Hence, there are 5 water molecules.

3 0

3 years ago

True or false?

Virty [35]

The answer would be true

3 0

3 years ago

What do these substances have in common?

Akimi4 [234]

They are compounds Good Luck!

4 0

2 years ago

Read 2 more answers

Freshwater is distributed in both time and space

Neko [114]

Answer:

Unevenly

Explanation:

Fresh water is distributed unevenly in both time and space.

7 0

1 year ago