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3 years ago

Which of the following gases would be most likely to experience ideal behavior at high pressures?

Chemistry

2 answers:

Delvig [45]3 years ago

4 0

D, <span>Monotonic gases, which have no inter molecular attractions are most suited as ideal gases </span><span />

Anuta_ua [19.1K]3 years ago

3 0

Answer:

The answer is D, just answered my test ;))))))))

Explanation:

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50.00 mL of 0.10 M HNO 2 (nitrous acid, K a = 4.5 × 10 −4) is titrated with a 0.10 M KOH solution. After 25.00 mL of the KOH sol

boyakko [2]

Answer:

b. 3.35

Explanation:

To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.

pH = pKa + log ([salt]/[acid]) (Eq. 01)

Where

pKa = -log(Ka) (Eq. 02)

[salt] = Molar concentration of salt produced as a result of titration

[acid] = Molar concentration of acid left in the solution after titration

Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:

HNO2 + KOH ⇆ H2O + KNO2 (Eq. 03)

This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.

Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles

Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles

As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.

Therefore

Amount of salt produced i.e [salt] = 0.0025 moles (Eq. 04)

Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)

Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:

pH= -log(4.5x10 -4) + log (0.0025/0.0025)

Solving above we get

pH = 3.35

5 0

3 years ago

Box 1

Alex73 [517]

10g

Explanation:

Box 1, Mass of A = 10g

Box 2, Mass of B = 5g

Box 3, = 1A + 1B

Unknown:

Mass of B that would combine with mass of 20g of A

Solution:

Mass ratio of A to B:

$\frac{mass of A}{mass of B}$ = mass ratio

$\frac{10}{5}$ = mass ratio

The mass ratio of A to B = 2: 1

Now, number of B that will combine with 20g of A;

$\frac{mass of A}{mass of B}$ = mass ratio

$\frac{20}{mass of B}$ = $\frac{2}{1}$

Mass of B = 10g

10g of B would combine with 20g of A

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7 0

3 years ago

Read 2 more answers

glucose 6‑phosphate+H2O⟶glucose+Pi glucose 6‑phosphate+H2O⟶glucose+Pi K′eq1=270 K′eq1=270 ATP+glucose⟶ADP+glucose 6‑phosphate AT

ddd [48]

Answer:

-30.7 kj/mol

Explanation:

The standard free energy for the given reaction that is the hydrolysis of ATP is calculated using the formula: ∆Go ’= -RTln K’eq

where,

R = -8.315 J / mo

T = 298 K

For reaction,

1. K′eq1=270,

∆Go ’= -RTln K’eq

= - 8.315 x 298 x ln 270

= - 8.315 x 298 x 5.59

= - 13,851.293 J / mo

= - 13.85 kj/mol

2. K′eq2=890

∆Go ’= -RTln K’eq

= - 8.315 x 298 x ln 890

= - 8.315 x 298 x 6.79

= - 16.82 kj/mol

therefore, total standard free energy

= - 13.85 + (-16.82)

= -30.7 kj/mol

Thus, -30.7 kj/mol is the correct answer.

6 0

3 years ago

What is the mass of solute in a 500 mL solution of 0.200 M

Fofino [41]

16.4 grams is the mass of solute in a 500 mL solution of 0.200 M .

sodium phosphate

Explanation:

Given data about sodium phosphate

atomic mass of Na3PO4 = 164 grams/mole

volume of the solution = 500 ml or 0.5 litres

molarity of sodium phosphate solution = 0.200 M

The formula for molarity will be used here to know the mass dissolved in the given volume of the solution:

The formula is

molarity = $\frac{number of moles of solute}{volume in litres}$

putting the values in the equation, we get

molarity x volume = number of moles

0.200 X 0.5= number of moles

number of moles = 0.1 moles

Atomic mass x number of moles = mass

putting the values in the above equation

164 x 0.1 = 16.4 grams

16.4 grams of sodium phosphate is present in 0.5 L of the solution to make a 0.2 M solution.

8 0

2 years ago

A 45 mL sample of nitrogen gas is cooled from 135ºC to 15C in a container that can contract or expand at constant pressure. Wha

Vanyuwa [196]

Answer:

V₂ =31.8 mL

Explanation:

Given data:

Initial volume of gas = 45 mL

Initial temperature = 135°C (135+273 =408 K)

Final temperature = 15°C (15+273 =288 K)

Final volume of gas = ?

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 = 45 mL × 288 K / 408 k

V₂ = 12960 mL.K / 408 K

V₂ =31.8 mL

8 0

2 years ago