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2 years ago

A weather balloon calibrated at 0.00 °C to

Chemistry

1 answer:

Natasha2012 [34]2 years ago

6 0

Answer:

17.3 L

Explanation:

With all other variables held constant, you can find the missing volume using the Charles' Law equation:

V₁ / T₁ = V₂ / T₂

In this equation, "V₁" and "T₁" represent the initial volume and temperature. "V₂" and "T₂" represent the final volume and temperature. Before we can solve, we need to convert Celsius to Kelvin (because we don't want negative numbers or temp. of 0).

V₁ = 20.0 L V₂ = ? L

T₁ = 0.00 °C + 273.15 = 273.15 K T₂ = -36.2 °C + 273.15 = 236.95 K

V₁ / T₁ = V₂ / T₂ <----- Charles' Law equation

20.0 L / 273.15 K = V₂ / 236.95 K <----- Insert values

0.0732 = V₂ / 236.95 K <----- Simplify left side

17.3 L = V₂ <----- Multiply both sides by 236.95

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Which type of statement provides a possible answer to a scientific question

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Explanation:

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2 years ago

Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +

qwelly [4]

<u>Answer:</u> The $\Delta H^o_{rxn}$ for the reaction is -1052.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)$ $\Delta H_1=-668.5kJ$

(2) $XCO_3(s)\rightarrow XO(s)+CO_2$ $\Delta H_2=+384.3kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ$

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3 years ago

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Explanation:

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2 years ago

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jolli1 [7]

Answer:

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