What pH does an acid have?

Complete combustion of 5.90 g of a hydrocarbon produced 19.2 g of CO2 and 5.89 g of H2O. What is the empirical formula for the h

scZoUnD [109]

$44g \ CO_{2} \ \ \ \rightarrow \ \ 12g \ C\\
19,2g \ CO_{2} \rightarrow \ \ \ x\\\\
x=\frac{19,2g*12g}{44g}\approx5,24g \ \ \ \Rightarrow \ \ n=\frac{5,24g}{12\frac{g}{mol}}\approx0,44mol\\\\\\
18g \ H_{2}O \ \ \ \rightarrow \ \ \ 2g \ H\\
5,89g \ H_{2}O \ \rightarrow \ \ y\\\\
y=\frac{5,89g*2g}{18g}\approx0,65g \ \ \ \ \Rightarrow \ \ n=\frac{0,65g}{1\frac{g}{mol}}=0,65mol\\\\\\
n_{C}:n_{H}=0,44:0,65\approx1:1\\\\
CH$

4 0

3 years ago

Which of the following statements is true

Morgarella [4.7K]

A. you're exposed to nuclear radiation everyday

6 0

3 years ago

What is the kinetic energy of a 1-kilogram ball is thrown into the air with ab ititial velocity of 30m/s?

mariarad [96]

KE=1/2*mass*velocity^2
So u do 1/2 * 1 * 30^2
1/2 * 1 * 900
= 450kgm/s

P.s. I'm not sure if I would have to convert kg to g.
Anyways hope this helped

4 0

3 years ago

Out of 500 silicon atoms: 460 are si-28, which has a mass of 27.98 amu 25 are si-29, which has a mass of 28.98 amu and 15 are si

liberstina [14]

Answer;

=28.09 amu

Explanation;

In this problem, they did not give us the percentages. However, since we know the number of atoms, we can easily calculate the percentages. For example:

(460 X 100)/500 = 92%

If we do this for all three isotopes,

(460 × 25)/500 = 5 %

(460 × 15) /500 = 3%

-We get 92%, 5%, and 3%. (We'll assume these are absolute numbers for determining our significant figures).

Now the problem is just like the previous one. First convert the percentages into decimals. Then multiply those decimals by the masses and add. Here's the solution:

= (0.92) X (27.98 amu) + (0.05) X (28.98 amu) + (0.03) X (29.97 amu)

= 25.74 amu + 1.449 amu + 0.8991 amu

= 28.09 amu

6 0

3 years ago

Read 2 more answers

Which of the following you is true for a limiting reactant

skad [1K]

Answer:

C) The limiting reactant has the lowest ratio of moles available / coefficient in the balanced equation.

Explanation:

Please, find attached a complete question to determine which of the statements is or are true for a limiting reactant in a chemical equation.

First, remember that the limiting reactant is the substance that is consumed completely while the excess reactant is the substance that does not react completely.

The limiting reactant is found comparing the stoichiometry ratio and the actual ratio between the reactants.

The stoichiometry ratio is found using the coefficientes of the chemical equation.

For illustration, assume the general chemical equation:

$aA+bB\rightarrow cC+dD$

The stoichiometric ratio of the reactants is:

$a\text{ }moles\text{ }of\text{ }A/b\text{ }moles\text{ }of\text{ }B$

If the ratio of the available moles of substance A to the available moles of substance B is greater than the stoichiometric ratio, it means that there are more moles of the substance A than what is needed to react with the available moles of substance B, then A will be in excess and B will B the limiting reactant.

If, on the contrary, the ratio of the available moles of substance A to the available moles of substance B is is less than the stoichiometric ratio, then substance A is less than the necessary to make the all the moles of substance B react, meaning that the substance A will limit the reaction (it will be consumed completely), while the substance B will be in excess.

As for the options:

A) The limiting reactant is has the lowest coefficient in a balanced equation.

This is false, since it is not the magnitude of the coefficiente what determines the limiting reactant, but the comparison of the ratios.

B) The limiting reactant is the reactant for which you have the fewest number of moles.

This is false because it is not the number of moles what determines the limiting reactant , but the comparison of the ratios.

C) The limiting reactant has the lowest ratio of moles available / coefficient in the balanced equation.

This is true as proved below.

The stoichiometric ratio of the reactants is:

$a\text{ }moles\text{ }of\text{ }A/b\text{ }moles\text{ }of\text{ }B$

The actual ratio is:

$available\text{ }moles\text{ }of\text{ }A/available\text{ }moles\text{ }of\text{ }B$

Assume the first ratio is less than the second (which describes when the substance A is in excess and the limiting reactant is the substance B).

$a\text{ }moles\text{ }of\text{ }A/b\text{ }moles\text{ }of\text{ }B$

Change the relation to show the ratios of moles available of each substance to the cofficient in the chemical equation:

$available\text{ }moles\text{ }of\text{ }B/b\text{ }moles\text{ }of\text{ }B$

Then, in the scenary that the limiting reactant is the substance B, the ratio of the left is lower than the ratio of the right, which is the same that limiting reactant has the lowest ratio of moles available / coefficient in the balanced equation.

D) The limiting reactant has the lowest ratio of coefficients in the balanced eqution/moles available.

This ratio is the inverse of the ratio of the previous statement, thus the relation is inverse, and, since the previous statement was true, this statement is false.

3 0

3 years ago