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3 years ago

Which of the four solutions is the best buffer against addition of acid or base?

Chemistry

1 answer:

Vera_Pavlovna [14]3 years ago

8 0

b. Na2HPO4 + NaH2PO4.

A <em>buffer </em>is a solution of a weak acid and its conjugate base. The weak acid is H2PO4^(-) and its conjugate base is HPO4^(2-).

All the other options are incorrect because they consist of only a single component.

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The atomic mass of a atom (in amu) is equal to

kobusy [5.1K]

1.67377x10-27 kilogram(kg) or 1.67377x 10-24 gram(g)

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3 years ago

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The zinc within a copper-plated penny will dissolve in hydrochloric acid if the copper coating is filed down in several spots (s

Mrac [35]

Answer:

Mass of hydrogen gas evolved is 0.0749 grams.

Explanation:

Total pressure of the gases = p = 758 mmHg

Vapor pressure of water = 23.78 mmHg

Pressure of hydrogen gas ,P = p - 23.78 mmHg = 758 mmHg - 23.78 mmHg

P = 734.22 mmHg = $\frac{734.22}{760} atm=0.966 atm$

Temperature of of hydrogen gas ,T= 25°C =298.15 K

Volume of hydrogen gas = V = 0.949 L

Moles of hydrogen gas =n

PV = nRT (Ideal gas equation )

$n=\frac{PV}{RT}=\frac{0.966 atm\times 0.949 L}{0.0821 atm L/mol K\times 298.15 K}$

n = 0.03745 mol

Moles of hydrogen gas = 0.03745 mol

Mass of 0.03745 moles of hydrogen gas = 0.03745 mol × 2 g/mol = 0.0749 g

Mass of hydrogen gas evolved is 0.0749 grams.

3 0

3 years ago

A 60.0 g block of iron that has an initial temperature of 250. °C and 60.0 g bloc of gold that has an initial temperature of 45.

Maslowich

Answer:

The final temperature at the equilibrium is 204.6 °C

Explanation:

Step 1: Data given

Mass of iron = 60.0 grams

Initial temperature = 250 °C

Mass of gold = 60.0 grams

Initial temperature of gold = 45.0 °C

The specific heat capacity of iron = 0.449 J/g•°C

The specific heat capacity of gold = 0.128 J/g•°C.

Step 2: Calculate the final temperature at the equilibrium

Heat lost = Heat gained

Qlost = -Qgained

Qiron = -Qgold

Q=m*c*ΔT

m(iron) * c(iron) *ΔT(iron) = -m(gold) * c(gold) *ΔT(gold)

⇒with m(iron) = the mass of iron = 60.0 grams

⇒with c(iron) = the specific heat of iron = 0.449 J/g°C

⇒with ΔT(iron)= the change of temperature of iron = T2 - T1 = T2 - 250.0°C

⇒with m(gold) = the mass of gold= 60.0 grams

⇒with c(gold) = the specific heat of gold = 0.128 J/g°C

⇒with ΔT(gold) = the change of temperature of gold = T2 - 45.0 °C

60.0 *0.449 * (T2 - 250.0) = -60.0 * 0.128 * (T2 - 45.0 )

26.94 * (T2 - 250.0) = -7.68 * (T2 - 45.0)

26.94T2 - 6735 = -7.68T2 + 345.6

34.62T2 = 7080.6

T2 = 204.5 °C

The final temperature at the equilibrium is 204.6 °C

5 0

3 years ago

How many liters are in 5.98 moles of nitrogen gas at STP?

Shtirlitz [24]

Answer:

The correct answer is c) 134L

Explanation:

We use the formula PV =nRT. The normal conditions of temperature and pressure are 273K and 1 atm, we use the gas constant = 0, 082 l atm / K mol.

1 atm x V = 5, 98 mol x 0, 082 l atm / K mol x 273 K

V = 5, 98 mol x 0, 082 l atm / K mol x 273 K / 1 atm

V = 133, 86828 l

8 0

3 years ago

What going on in each spot

KATRIN_1 [288]

nothing ksbsshshhzvsjajbsjshjsgdvdjhsbsj

3 0

3 years ago

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