Which of the following is true about DNA?

A student calculated the molarity of a solution prepared by dissolving 0.730 mol of table sugar (sucrose, C12H22O11) in 1.8x10^3

lara31 [8.8K]

Answer: C= 0.406 M

Explanation:

Solution.

ν

=

0.730

m

o

l

;

ν=0.730mol;

V

=

1.8

⋅

1

0

3

m

L

=

1.8

L

;

V=1.8⋅10

3 mL=1.8L;

C=0.730mol

1.8 L=0.406 M

C= 1.8L

0.730mol =0.406M

The student made a mistake because he did not convert a unit of volume from milliliters to liters. After all, molarity is defined as the number of moles of solute per liter of solution.

5 0

3 years ago

1.Which Statement best describes the use of a catalyst in a fuel cell?

mixer [17]

Answer: D. It Increases reaction rates.

Explanation:

4 0

2 years ago

Read 2 more answers

In a strong acid–strong base titration (both monoprotic), if 25.0 milliliters of the base is required to completely neutralize 2

taurus [48]

B)The concentration of the acid (C₁) is the same as that of the base (C₂).

V₁=V₂

n(acid)=C₁V₁
n(base)=C₂V₂

HX + YOH = YX + H₂O

n(acid)=n(base)

C₁V₁=C₂V₂

C₁=C₂

5 0

3 years ago

Read 2 more answers

Problem Page Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 70. g of

shusha [124]

Answer:

The maximum mass of carbon dioxide that could be produced by the chemical reaction is 70.6gCO_{2}

Explanation:

1. Write down the balanced chemical reaction:

$2C_{6}H_{14}_{(l)}+19O_{2}_{(g)}=12CO_{2}_{(g)}+14H_{2}O_{(g)}$

2. Find the limiting reagent:

- First calculate the number of moles of hexane and oxygen with the mass given by the problem.

For the hexane:

$70.0gC_{6}H_{14}*\frac{1molC_{6}H_{14}}{86.2gC_{6}H_{14}}=0.81molesC_{6}H_{14}$

For the oxygen:

$81.3gO_{2}*\frac{1molO_{2}}{32.0gO_{2}}=2.54molesO_{2}$

- Then divide the number of moles between the stoichiometric coefficient:

For the hexane:

$\frac{0.81}{2}=0.41$

For the oxygen:

$\frac{2.54}{19}=0.13$

- As the fraction for the oxygen is the smallest, the oxygen is the limiting reagent.

3. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction:

The calculations must be done with the limiting reagent, that is the oxygen.

$81.3gO_{2}*\frac{1molO_{2}}{32gO_{2}}*\frac{12molesCO_{2}}{19molesO_{2}}*\frac{44.0gCO_{2}}{1molCO_{2}}=70.6gCO_{2}$

7 0

3 years ago

A. 14.8

Scilla [17]

Answer:

Explanation:

1)

Given data:

Initial volume = 2.5 L

Initial temperature = 300 k

Final temperature = 80 k

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 2.5 L × 80 K / 300 k

V₂ = 200 L.K / 300 K

V₂ = 0.67 L

2)

Given data:

Initial volume = 752 mL

Initial temperature = 25.0°C (25+273 = 298 K)

Final temperature = 50.0°C(50+273 = 323 K)

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 752 mL × 323 K / 298 k

V₂ = 242896 mL.K / 298 K

V₂ = 815.1 mL

3)

Given data:

Initial volume = 2.75 L

Initial temperature = 20°C (20+273 = 293 K)

Final temperature = ?

Final volume = 2.46 L

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁/V₁

T₂ = 2.46 L × 293 K / 2.75 L

T₂ = 720.78 L.K / 2.75 L

T₂ = 262.1 K

4)

Given data:

Initial volume = 1500 L

Initial temperature = 5°C (5+273 = 278 K)

Final temperature = 30 °C(30+273 = 303 K)

Final volume of heated air = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 1500 L × 303 K / 278 k

V₂ = 454500 L.K / 278 K

V₂ = 1634.89 L

5)

Given data:

Initial volume = 15.5 L

Initial temperature = 20°C (20+273 = 293 K)

Final temperature = 7.0 °C(7.0+273 = 280 K)

Final volume of balloon = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 15.5 L × 280 K / 293 k

V₂ = 4340 L.K / 293 K

V₂ = 14.8 L

6)

Given data:

Initial volume = 150 mL

Initial temperature = 23.5°C (23.5+273 = 296.5 K)

Final temperature = 72.5 °C(72.5+273 = 345.5 K)

Final volume of balloon = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 150 mL × 345.5 K / 296.5 k

V₂ = 51825 mL.K / 296.5 K

V₂ = 174.79 mL

6 0

3 years ago