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3 years ago

1. A 4.00 L sample of air at 35 degrees C expands to 5.50 L when heated. What is

Chemistry

1 answer:

tatuchka [14]3 years ago

3 0

Answer:

150.5 °C

Explanation:

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

Initial volume (V₁) = 4 L

Initial temperature (T₁) = 35 °C

Final volume (V₂) = 5.5 L

Final temperature (T₂) =?

Next, we shall convert 35 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 35 °C

Initial temperature (T₁) = 35 °C + 273

Initial temperature (T₁) = 308 K

Next, we shall determine the final (i.e the new) temperature of air. This can be obtained as follow:

Initial volume (V₁) = 4 L

Initial temperature (T₁) = 308 K

Final volume (V₂) = 5.5 L

Final temperature (T₂) =?

V₁ / T₁ = V₂ / T₂

4 / 308 = 5.5 / T₂

Cross multiply

4 × T₂ = 308 × 5.5

4 × T₂ = 1694

Divide both side by 4

T₂ = 1694 / 4

T₂ = 423.5 K

Finally, we shall convert 423.5 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

T₂ = 423.5 K

T₂ = 423.5 – 273

T₂ = 150.5 °C

Thus, the new temperature of the air is 150.5 °C

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balu736 [363]

The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is
C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is
6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)

Using the equation for the standard enthalpy change of formation
ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}

C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
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3 years ago

If you are given a 0. 31 g piece of sodium metal to react with water, how many moles of hcl would it take to neutralize the sodi

Sphinxa [80]

The moles of HCl to neutralize the sodium hydroxide produced is 0.0135 mole.

Neutralization or neutralization is a chemical response wherein an acid and a base react quantitatively with each other. In a reaction in water, neutralization outcomes in there being no excess of hydrogen or hydroxide ions gift in the answer.

calculation:-

2Na + 2H₂O -----> 2NaOH + H₂

2 mol or 46g of Na produces 80 grams of NaOH

∴ 0.31 g of Na will produce = 80/46 × 0.31

= 0.54 gram of NaOH.

mol of NaOH = 0.54/40

= 0.0135

Since both Hcl and NaOH have the same valance factor,

1 mole NaOH is needed to neutralize 1 mol HCl

∴ 0.0135 mole of NaOH will need = 0.0135 mole of HCl

mass = 0.0135 × 36.5

= 0.493 grams of HCL.

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1 year ago

Calculate the number of moles CuCl2 necessary to make 50 mL of a 0.15M solution

tekilochka [14]

Volume ⇒ 50 mL in liters : 50 / 1000 = 0.05 L

Molarity of solution ⇒ 0.15 M

Number of moles:

n = M * V

n = 0.15 * 0.05

n = 0.0075 moles of CuCl2

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3 years ago

Prandtl number in heat transfer analogues to which dimension less number in mass transfer. a) Schmidt number b) Sherwood number

Cloud [144]

Answer:

a) Schmidt number

Explanation:

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