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3 years ago

Five liters of air at -35°C is warmed to 89°C. What is the new volume if the pressure remains constant

Chemistry

1 answer:

Delicious77 [7]3 years ago

8 0

Answer:

7.61 liters

Explanation:

From the question,

Applying Charles law formula,

V₁/T₁ = V₂/T₂...................... Equation 1

Where P₁ = initial volume of of air, V₂ = Final volume of air, T₁ = Initial Temperature of air, T₂ = Final Temperature of air.

make V₂ the subject of the equation

V₂ = (V₁×T₂)/T₁............... Equation 2

Given: V₁ = 5 liters, T₁ -35 °C = (-35+273) K = 238 K, T₂ = 89°C = (89+273) K = 362 K

Substitute these values into equation 2

V₂ = (5×362)238

V₂ = 1810/238

V₂ = 7.61 liters

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Dvinal [7]

Answer:

2.81 × 10⁶ mm³

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Explanation:

Step 1: Given data

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Thickness (t): 50 mm

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3 years ago

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Answer:

$m=59.3gCO_2$

Explanation:

Hello,

In this case, the first step is to compute the molar mass of carbon dioxide as shown below, considering it has one carbon atom and two oxygen atoms:

$M=12.011g/mol+2*15.999g/mol\\\\M=44.009g/mol$

It is important to notice it is the mass in one mole of such compound. Afterwards, we need to use the Avogadro's number to compute the how many moles are in the given molecules of carbon dioxide as shown below:

$mol=8.12x10^{23}molec*\frac{1mol}{6.022x10^{23}molec} =1.35mol$

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4 years ago

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The decomposition of NI3 to form N2 and I2 releases −290.0 kJ of energy. The reaction can be represented as 2NI3(s)→N2(g)+3I2(g)

EastWind [94]

Answer:

-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.

Explanation:

Mass of nitrogen triiodide = 20.0 g

Moles of nitrogen triiodide = $\frac{20.0 g}{395 g/mol}=0.05063 mol$

$2NI_3(s)\rightarrow N_2(g)+3I_2(g), \Delta H_{rxn}=-290.0 kJ$

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