A solution is classified as a weak base. Which of these could be the pH of the solution?

K = °C + 273 A 4.1 L sample of gas is held at 25 °C. If the gas expands to 6.8 L, what is the final temperature?

gregori [183]

Answer:

221 °C

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 4.1 L

Initial temperature (T₁) = 25 °C

= 25 °C + 273

= 298 K

Final volume (V₂) = 6.8 L

Final temperature (T₂) =?

The final temperature of the gas can be obtained as follow:

V₁ / T₁ = V₂ / T₂

4.1 / 298 = 6.8 / T₂

Cross multiply

4.1 × T₂ = 298 × 6.8

4.1 × T₂ = 2026.4

Divide both side by 4.1

T₂ = 2026.4 / 4.1

T₂ ≈ 494 K

Finally, we shall convert 494 K to celcius temperature. This can be obtained as follow:

°C = K – 273

K = 494

°C = 494 – 273

°C = 221 °C

Thus the final temperature of the gas is 221 °C

6 0

3 years ago

Sodium permanganate and iron(III) chloride are both dissolved in a beaker of water. In a double displacement reaction, two new p

Nastasia [14]

<u>Answer:</u> The balanced chemical equation is written below.

<u>Explanation:</u>

Double displacement reaction is defined as the reaction in which exchange of ions takes place.

$AB+CD\rightarrow CB+AD$

When sodium permanganate reacts with iron (III) chloride, it leads to the production of sodium chloride and iron (III) permanganate.

The chemical equation for the reaction of sodium permanganate and iron (III) chloride follows:

$3NaMnO_4(aq.)+FeCl_3(aq.)\rightarrow Fe(MnO_4)_3(s)+3NaCl(aq.)$

By Stoichiometry of the reaction:

3 moles of aqueous solution of sodium permanganate reacts with 1 mole of aqueous solution of iron (III) chloride to produce 1 mole of solid iron permanganate and 3 moles of aqueous solution of sodium chloride

Hence, the balanced chemical equation is written above.

7 0

3 years ago

A fixed amount of gas at 25.0°C occupies a volume of 10.0 L when the pressure is 667 torr. Use Boyle's law to calculate the pres

zaharov [31]

Answer:

The pressure, when the volume is reduced to 7.88L, is 846 torr (option A)

Explanation:

Step 1: Data given

The temperature of a gas = 25.0°C

AT 25 °C the gas occupies a volume of 10.0L and a pressure of 667 torr.

The volume reduces to 7.88 L but the temperature stays constant.

Step 2: Boyle's law

(P1*V1)/T1 = (P2*V2)/T2

⇒ Since the temperature stays constant, we can simplify to:

P1*V1 = P2*V2

⇒ with P1 = the initial pressure 667 torr

⇒ with V1 = the initial volume = 10.0 L

⇒ with P2 = the final pressure = TO BE DETERMINED

⇒ with V2 = the final volume = 7.88L

P2 = (P1*V1)/V2

P2 = (667*10.0)/7.88

P2 = 846 torr

The pressure, when the volume is reduced to 7.88L, is 846 torr (option A)

6 0

2 years ago

What evidence supports conservation of matter?

Zinaida [17]

Answer:

i dont knoq

Explanation:

sorry that i couldnt help you

7 0

2 years ago

The solubility of glucose at 30°C is

weqwewe [10]

Answer:

Saturated solution

We should raise the temperature to increase the amount of glucose in the solution without adding more glucose.

Explanation:

Step 1: Calculate the mass of water

The density of water at 30°C is 0.996 g/mL. We use this data to calculate the mass corresponding to 400 mL.

$400 mL \times \frac{0.996g}{1mL} =398g$

Step 2: Calculate the mass of glucose per 100 g of water

550 g of glucose were added to 398 g of water. Let's calculate the mass of glucose per 100 g of water.

$100gH_2O \times \frac{550gGlucose}{398gH_2O} = 138 gGlucose$

Step 3: Classify the solution

The solubility represents the maximum amount of solute that can be dissolved per 100 g of water. Since the solubility of glucose is 125 g Glucose/100 g of water and we attempt to dissolve 138 g of Glucose/100 g of water, some of the Glucose will not be dissolved. The solution will have the maximum amount of solute possible so it would be saturated. We could increase the amount of glucose in the solution by raising the temperature to increase the solubility of glucose in water.

6 0

3 years ago