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3 years ago

15 g de nitrato de amonio, NH4NO3 se disuelven en 135 g de agua. La densidad de la

Chemistry

1 answer:

MrRa [10]3 years ago

7 0

Answer:

δ sln = 0.111 g/mL

Explanation:

density (δ): mass/volume

⇒ δ ≡ g solute/mL H2O

∴ m NH4NO3 = 15g

∴ δH2O ≅ 1 g/mL

∴ V H2O = (135g H2O).( mL/1gH2O) = 135 mL H2O

⇒ δ sln = 15 g NH4NO3 / 135 mL H2O = 0.111 g/mL

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F. How many centigrams are in 253,000 picograms? Plz show work

Darina [25.2K]

The answer is 2.53e-5, I unfortunately don't know how you would really show the work other than showing the division.

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3 years ago

Solid-solid homogeneous mixture give one example

likoan [24]

Example of solid - solid homogeneous mixture is copper metal - silver metal like coins and alloys. Homogeneous mixture is a mixture in which one of the substances often changes in form as in a solution of sugar in water. It contains variable proportions. Solution can contain two substances, three substances or more, in a single physical state. The component of a solution that is present in greatest quantity is usually called the solvent and all other components are called solutes.

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3 years ago

Read 2 more answers

Why should you always condition a buret before running a titration?.

Alex Ar [27]

Answer:

Conditioning two or three times will insure that the concentration of titrant is not changed by a stray drop of water.

Explanation:

"Check the tip of the buret for an air bubble. To remove an air bubble, whack the side of the buret tip while solution is flowing".

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2 years ago

While heating up a 25 gram sample of concrete (specific heat = 0.210-cal/g°C), your initial tempărature is room temperature (25°

Lana71 [14]

Answer:

Final temperature = 83.1 °C

Explanation:

Given data:

Mass of concrete = 25 g

Specific heat capacity = 0.210 cal/g. °C

Initial temperature = 25°C

Calories gain = 305 cal

Final temperature = ?

Solution:

Q = m. c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

305 cal = 25 g ×0.210 cal/g.°C × T2 - 25°C

305 cal = 5.25cal/°C × T2 - 25°C

305 cal / 5.25cal/°C = T2 - 25°C

58.1 °C = T2 - 25°C

T2 = 58.1 °C + 25°C

T2 = 83.1 °C

7 0

2 years ago

A buffer contains 0.18 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. What is the pH

Yuki888 [10]

Answer:

1) pH = 5.05

2) pH = 5.13

3) pH = 4.97

Explanation:

Step 1: Data given

Number of moles of propionic acid = 0.18 moles

Number of moles sodium propionate = 0.26 moles

Volume = 1.20 L

Ka = 1.3 * 10^-5 → pKa = 4.989

Step 2: Calculate concentrations

Concentration = moles / volume

[acid]= 0.18/ 1.2 =0.150 M

[salt]= 0.26/ 1.3 = 0.217 M

pH = 4.89 + log(0.217/0.150)=5.05

What is the pH of the buffer after the addition of 0.02 mol of NaOH?

moles acid = 0.18 - 0.02 = 0.16

[acid]= 0.16/ 1.2=0.133 M

moles salt = 0.26 + 0.02 = 0.28

[salt]= 0.28/ 12=0.233

pH = 4.89 + log 0.233/ 0.133 = 5.13

What is the pH of the buffer after the addition of 0.02 mol of HI?

moles acid = 0.18+ 0.02 = 0.20 moles

[acid]= 0.20/ 1.2 = 0.167 M

[salt]= 0.26 - 0.02= 0.24 moles

[salt]= 0.24/ 1.2 = 0.20 M

pH = 4.89 + log 0.20/ 0.167= 4.97

8 0

3 years ago