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3 years ago

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Air pressure decreases more rapidly with altitude within a column of air than in a column of __ air.

1 answer:

Anarel [89]3 years ago

4 0

Air Pressure drops more rapidly with altitude in a column of cold air than in warm air.The answers to this question are cold air and warm air, respectively.
<span>Cold air is known to be dense while warm air is known otherwise to be less thens which makes it move upwards. Cold air experiences more pressure as it moves upwards.</span>

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Percentage composition

TiliK225 [7]

There goes the ist of what you're supposed to do.

3 0

3 years ago

Read 2 more answers

Calculate ΔHrxnΔHrxn for the following reaction: 3C(s)+4H2(g)→C3H8(l)3C(s)+4H2(g)→C3H8(l) Use the following reactions and given

mihalych1998 [28]

<u>Answer:</u> The $\Delta H^o_{rxn}$ for the reaction is -120.9 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:

$3C(s)+4H_2(g)\rightarrow C_3H_8(l)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $C_3H_8(l)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)$ $\Delta H_1=-2026.6kJ$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ$ ( × 3)

(3) $2H_2(g)+O_2(g)\rightarrow 2H_2O(g)$ $\Delta H_2=-483.5kJ$ ( × 2)

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (-\Delta H_1)]+[3\times \Delta H_2]+[2\times \Delta H_3]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times -(-2026.6))+(3\times (-393.5))+(2\times (-483.5))]=-120.9kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -120.9 kJ.

8 0

3 years ago

Could you please Calculate the number of atom of 40K (potassium 40) in 1gram of KCl. Taking into account the isotopic abundance

Crank

Answer:

$9.53*10^{17}$ atoms of 40K

Explanation:

You can use the molecular mass and the Avogadro´s number, in the following formula:

$N_{40K=\frac{m_{KCl}}{M_{MKCl}}}*N_{Avogadro}*(IA_{40k})$

where $m_{KCl}$ is the sample mass, $M_{KCl}$ is the molecular mass of the KCl and IA(40K) is the isotopic abundance of 40K.

Now replacing the values, you can find:

$N_{40K}=\frac{1g}{74.5513\frac{g}{mol}}*6.022*10^{23}mol^{-1} *0.000118$

$N_{40K}=9.53*10^{17} atoms$

8 0

3 years ago

Write a balanced chemical equation for the fermentation sucrose (C12H22O11) by yeasts in which the aqueous ethyl alcohol (C2H5OH

kodGreya [7K]

Answer:

C12H22O11(aq) + H2O(l) —> 4C2H5OH(aq) + 4CO2(g)

Explanation:

When aqueous sugar (sucrose) react with water in the presence of yeast, the following products are obtained as shown in the equation below:

C12H22O11(aq) + H2O(l) —> C2H5OH(aq) + CO2(g)

Now, we shall balance the equation as follow:

There are a total of 24 atoms of H on the left side and 6 atoms on the right side. It can be balance by putting 4 in front of C2H5OH as shown below:

C12H22O11(aq) + H2O(l) —> 4C2H5OH(aq) + CO2(g)

There are a total of 9 atoms of C on the right side and 12 atoms on the left side. It can be balance by putting 4 in front of CO2 as shown below:

C12H22O11(aq) + H2O(l) —> 4C2H5OH(aq) + 4CO2(g)

Now the equation is balanced.

8 0

3 years ago

A weak base. B. has a K. of 4.46 x 10-10 A solution with an unknown initial concentration is tested, and found to have a pH of 8

Ann [662]

Answer:

0.013MB

Explanation:

6 0

3 years ago