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2 years ago

List the number of each type of atom on the left side of the equation 2C10H22(l)+31O2(g)→20CO2(g)+22H2O(g)

Chemistry

1 answer:

Leokris [45]2 years ago

3 0

Answer:

20, 44, 62

Explanation:

To find the number of atoms of each element, we multiply coefficient and subscript

For example $5 Ca_1 Cl_2$ contains

5 × 1 = 5 ,Ca atoms and

5 × 2 = 10, Cl atoms

If there is a bracket in the chemical formula

For example $3Ca_3 (P_1 O_4 )_2$

we multiply coefficient × subscript × number outside the bracket to find the number of atoms

(Please note: 3 is the coefficient, and if there is no number given then 1 will be the coefficient )

3 × 3 = 9 , Ca atoms

3 × 1 × 2 = 6, P atoms

3 × 4 × 2 = 24, O atoms are present.

So let us find the number of atoms of each element on the left side of the equation

$2C_{10} H_{22} (l)+31O_2 (g)\Rightarrow 20CO_2 (g)+22H_2 O(g)$

Number of C atoms = 2 × 10 = 20

Number of H atoms = 2 × 22 = 44

Number of O atoms = 31 × 2 = 62

20, 44, 62 are the Answers.

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Answer:

34.02 g.

Explanation:

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In this case, since the gas behaves ideally, we can use the following equation to compute the moles at the specified conditions:

$PV=nRT\\\\n=\frac{1.00atm*0.960L}{0.08206\frac{atm*L}{mol*K}*(150+273)K} =0.0277mol\\\\$

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3 years ago

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2 years ago

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The reaction is endothermic

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3 years ago

In order to prepare 50.0 mL of 0.100 M NaOH you will add _____ mL of 1.00 M NaOH to _____ mL of water

FinnZ [79.3K]

The question requires us to complete the sentence regarding the preparation of a more dilute NaOH solution (0.100 M, 50.0 mL) from a more concentrated NaOH solution (1.00 M).

Analyzing the blank spaces that we need to fill in the sentence, we can see that we must provide the volume of the more concentrated solution and the volume of water necessary to prepare the solution.

We can use the following equation to calculate the volume of more concentrated solution required:

$\begin{gathered} C_1\times V_1=C_2\times V_2 \\ V_1=\frac{C_2\times V_2}{C_1} \end{gathered}$

where C1 is the concentration of the initial solution (C1 = 1.00 M), V1 is the volume required of the inital solution (that we'll calculate), C2 is the concentration of the final solution (C2 = 0.100 M) and V2 is the volume of the final solution (V2 = 50.0 mL).

Applying the values given by the question to the equation above, we'll have:

$\begin{gathered} V_1=\frac{C_2\times V_2}{C_1} \\ V_1=\frac{0.100M_{}\times50.0mL_{}}{1.00M_{}}=5.00mL \end{gathered}$

Thus, we would need 5.00 mL of the more concentrated solution.

Since the volume of the final solution is 50.0 mL and it corresponds to the volume of initial solution + volume of water, we can calculate the volume of water necessary as:

$\begin{gathered} \text{final volume = volume of initial solution + volume of water} \\ 50.0mL=5.00mL\text{ + volume of water} \\ \text{volume of water = 45.0 mL} \end{gathered}$

Thus, we would need 45.0 mL of water to prepare the solution.

Therefore, we can complete the sentence given as:

"In order to prepare 50.0 mL of 0.100 M NaOH you will add 5.00 mL of 1.00 M NaOH to 45.0 mL of water"

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11 months ago