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4 years ago

8

The percentage of water vapor present in the air compared to that required for saturation is the ____. a. mixing ratio b. relati

ve humidity c. dew point d. absolute humidity

1 answer:

Ivan4 years ago

7 0

Answer:

B. Relative Humidity

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Determine the limiting reactant when 30.0 g of propane, C3H8, is burned with 75.0 g of oxygen.

AnnZ [28]

<u>Answer:</u> The limiting reagent is oxygen gas.

<u>Explanation:</u>

Limiting reagent is defined as the reactant that is present in less amount and it limits the formation of products.

Excess reagent is defined as the reactant which is present in large amount.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For propane:</u>

Given mass of propane = 30.0 g

Molar mass of propane = 44.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of propane}=\frac{30.0g}{44.1g/mol}=0.680mol$

<u>For oxygen:</u>

Given mass of oxygen = 75.0 g

Molar mass of oxygen = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of propane}=\frac{75.0g}{32g/mol}=2.34mol$

The chemical equation for the combustion of propane follows:

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)$

By Stoichiometry of the reaction:

5 moles of oxygen gas reacts with 1 mole of propane.

So, 2.34 moles of oxygen gas will react with = $\frac{1}{5}\times 2.34=0.468mol$ of propane

As, given amount of propane is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reagent is oxygen gas.

3 0

3 years ago

How is beryllium (Be) similar to calcium (Ca)?

timofeeve [1]

C both are very reactive i guess

7 0

4 years ago

Given the equilibrium constants for the following two reactions in aqueous solution at 25 ∘C HNO2(aq)H2SO3(aq)⇌⇌H+(aq) + NO2−(aq

krok68 [10]

Answer : The value of $K_c$ for the final reaction is, 184.09

Explanation :

The equilibrium reactions in aqueous solution are :

(1) $HNO_2(aq)\rightleftharpoons H^+(aq)+NO_2^-(aq)$ $K_{c_1}=4.5\times 10^{-4}$

(2) $H_2SO_3(aq)\rightleftharpoons 2H^+(aq)+SO_3^{2-}(aq)$ $K_{c_2}=1.1\times 10^{-9}$

The final equilibrium reaction is :

$2HNO_2(aq)+SO_3^{2-}(aq)\rightleftharpoons H_2SO_3(aq)+2NO_2^-(aq)$ $K_{c}=?$

Now we have to calculate the value of $K_c$ for the final reaction.

Now equation 1 is multiply by 2 and reverse the equation 2, we get the value of final equilibrium reaction and the expression of final equilibrium constant is:

$K_c=\frac{(K_{c_1})^2}{K_{c_2}}$

Now put all the given values in this expression, we get :

$K_c=\frac{(4.5\times 10^{-4})^2}{1.1\times 10^{-9}}=184.09$

Therefore, the value of $K_c$ for the final reaction is, 184.09

4 0

3 years ago

A chemistry student needs 40.0 mL of pentane for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the stu

Alenkinab [10]

Answer:

25.04 g is the mass of pentane

Explanation:

Pentane density = Pentane mass / Pentane volume

1 cm³ = 1mL

0.626 g/mL = Pentane mass / Pentane volume

0.626 g/mL = Pentane mass / 40 mL

0.626 g/mL . 40 mL = Pentane mass → 25.04 g

5 0

3 years ago

Why does Beryllium have a larger first Ionization energy than boron?

Nata [24]

This is an exception to the general electronegativity trend. It can be explained by looking at the electron configurations of both elements.

<span>Be:[He]2<span>s2
</span></span><span>B:[He]2<span>s2</span>2<span>p1

</span></span>

When you remove an electron from beryllium, you are taking away an electron from the 2s orbital. When you remove an electron from boron, you are taking an electron from the 2p orbital. The 2p electrons have more energy than the 2s, so it is easier to remove them as they can more strongly resist the effective nuclear charge of the nucleus.

7 0

4 years ago