Which substance has a giant covalent structure and contains atoms of more than one element?A diamondB graphiteC methaneD sand

Will mark brainliest if correct

Sergeeva-Olga [200]

Answer:

a and d

Explanation:

4 0

2 years ago

You have a flask with 1.2 L of liquid in it, how many mL is this?

mihalych1998 [28]

I literally is 1000 ml. Therefor 1.2 literally is 1.2*1000 which is 1200 ml

6 0

3 years ago

Read 2 more answers

If 4.50 g of HCl are reacted with 15.00 g of Caco, according to the following balanced chemical equation, calculate the theoreti

Tom [10]

Answer: HCl is the limiting reactant and the theoretical yield is 2.72 g of CO2. If the actual yield was 2.50 g then, the percent yield is 92.0% when rounding off is done only for the final answer.

Further Explanation:

In order to determine the theoretical yield and the percent yield of CO2, the following steps must be done:

Determine the limiting reactant. This is the reactant that will determine the amount of CO2 that will actually form.
Determine the theoretical yield for CO2 when the limiting reactant is used.
Get the percent yield by getting the ratio of the actual yield stated in the problem and the calculated theoretical yield multiplied by 100.

Determining the Limiting Reactant

The Limiting Reactant (LR) will produce fewer moles of the products. To check which of the reactants HCl or CaCO3 is the LR, we do dimensional analysis:

For HCl:

$moles\ CO_{2}\ = (4.50\ g\ HCl)\(\frac{1\ mol\ HCl}{36.46094\ g})( \frac{1\ mol\ CO_{2} }{2\ mol\ HCl}) \\moles\ CO_{2}\ =\ 0.0617098$

For CaCO3:

$moles\ of\ CO_{2}\ = (15.00\ g\ CaCO_{3})\ (\frac{1\ mol\ CaCO_{3} }{100.0869\ g\ CaCO_{3} })\ (\frac{1\ mol\ CO_{2} }{1\ mol\ CaCO_{3} })\\moles\ of\ CO_{2}\ = \ 0.1499$

Since HCl produces fewer moles of CO2, then it is the limiting reactant. We will use the given amount to determine the theoretical yield for CO2.

Determining the Theoretical Yield

From Step 1, we know that 0.0617098 moles of CO2 will be produced. We will just convert this to grams.

$grams\ CO_{2}\ =\ (0.0617098\ mol\ CO_{2}) (\frac{44.01\ g\ CO_{2}}{1\ mol\ CO_{2}})\\grams\ CO_{2}\ =\ 2.71585$

Since the answer only requires 3 significant figures, the final answer is 2.72 grams CO2.

Determining the Percent Yield

Dividing the actual yield by the theoretical yield will give us the percent yield, which is an indicator of how efficient the experiment or the method used was.

From the problem, the actual yield was 2.50 g, hence, the percent yield is:

$percent\ yield\ of\ CO_{2}\ = (\frac{2.50\ g}{2.71585\ g}) (100)\\percent\ yield\ of\ CO_{2}\ = 92.05221$