Which of the following steps in a star's life cycle takes the longest?

What is the correct formula for selenium hexachloride?

labwork [276]

Answer:

Selenium hexachloride is SeCl6.

Explanation:

3 0

2 years ago

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Fill in the left side of this equilibrium constant equation for the reaction of benzoic acid with water

Vera_Pavlovna [14]

Answer:

8 0

3 years ago

A brown dye has a percent composition of 62.41% C, 5.24% H, and 32.36% N by mass with a molar mass of 346.40 g/mol. Determine th

Fiesta28 [93]

The molecular formula : C₁₈H₁₈N₈

<h3>Further explanation</h3>

Given

62.41% C, 5.24% H, and 32.36% N

Required

The molecular formula

Solution

mol ratio

C : 62.41/12.0096 = 5.1967

H : 5.24/1.00784 = 5.1992

N : 32.36/14.0067 = 2.310

Divide by 2.310(smallest)

C : 5.1967/2.31=2.25

H : 5.1992/2.31 = 2.25

N : 2.31/2.31 = 1

Multiplied by 4

C : H : N = 9 : 9 : 4

The empirical formula : C₉H₉N₄

(C₉H₉N₄)n=346.40 g/mol

(12.0096 x 9 + 1.00784 x 9 + 14.0067 x 4)n=346.4

(108.0864+9.07056+56.0268)n=346.4

(173.184)n=346.4

n=2

<em>The molecular formula : C₁₈H₁₈N₈</em>

8 0

2 years ago

Based on the information, what is the balanced chemical equation representing the burning of butane?

kompoz [17]

Answer:

OptionA. 2C4H10 + 13O2 —> 8CO2 + 10H20

Explanation:

Butane burns is air (O2) according to the equation:

C4H10 + O2 —> CO2 + H20

Considering the equation, it is evident that it not balanced. Now let us balance the equation as shown below;

There are a total of 4 carbon atoms on the left and 1 carbon atom on the right. It can be balanced by putting 4 in front of CO2 as shown below:

C4H10 + O2 —> 4CO2 + H20

Next, there are 10 hydrogen atoms on the left and 2 hydrogen atoms on the right. Therefore to balance it, put 5 in front of H2O as shown below:

C4H10 + O2 —> 4CO2 + 5H20

Now, there are a total of 13 oxygen atoms on the right and 2 at the left. To balance it, put 13/2 in front of O2

as shown below

C4H10 + 13/2O2 —> 4CO2 + 5H20

Now we multiply through by 2 clear off the fraction and we obtained:

2C4H10 + 13O2 —> 8CO2 + 10H20

6 0

3 years ago

How many grams of lead ii sulfide is produced when 25.0 g lead ii acetate reacts with excess hydrogen sulfide

ddd [48]

Answer: 18.42 grams of lead (II) sulfide will be produced in the given reaction:

Explanation: The reaction of lead (II) acetate and hydrogen sulfide follows:

$(CH_3COO)_2Pb+H_2S\rightarrow PbS+2CH_3COOH$

To calculate the moles, we use the formula:

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

Molar mass of lead (II) acetate = 325.29 g/mol

Given mass of lead (II) acetate = 25 g

Putting values in above equation, we get:

$Moles=\frac{25g}{325.29g/mol}=0.0768moles$

We are given that hydrogen sulfide is present in excess, so limiting reagent is lead (II) acetate because it limits the formation of product.

By stoichiometry of the reaction,

1 moles of lead (II) acetate produces 1 mole of lead (II) sulfide

So, 0.0768 moles of lead (II) acetate will produce = $\frac{1}{1}\times 0.0768$ = 0.0768 moles of lead (II) sulfide.

Now, to calculate the mass of lead (II) sulfide, we use equation 1, we get:

Molar mass of lead (II) sulfide = 239.3 g/mol

$0.0768mol=\frac{\text{Given mass}}{239.3g/mol}$

$\text{Mass of lead (II) sulfide}=18.42g$

8 0

3 years ago