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Hunter-Best [27]
2 years ago
5

Which statement best describes the effects chemicals may have on the

Chemistry
2 answers:
frutty [35]2 years ago
7 0
The answer is B. So yeah
FromTheMoon [43]2 years ago
4 0
Some chemicals may help the environment
You might be interested in
According to the Law of Conservation of Mass, the total
dimulka [17.4K]

Answer:

Always equal to the total moles of the products.

Explanation:

The law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to the law of conservation of mass, the mass of the products in a chemical reaction equal to the mass of the reactants.

8 0
2 years ago
The only state in which matter has a fixed shape is ___?
qwelly [4]

Answer:

solid

Explanation:

Matter in the solid state maintains a fixed volume and shape

5 0
3 years ago
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:

  1. Determine the limiting reactant. This is the reactant that will determine the amount of CO2 that will actually form.
  2. Determine the theoretical yield for CO2 when the limiting reactant is used.
  3. 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

Rounding off to three significant figures, the percent yield is 92.0%. This suggests that the method used is somewhat efficient in producing CO2.

Learn More

  1. Learn More about Limiting Reactant brainly.com/question/7144022
  2. Learn More about Excess Reactant brainly.com/question/6091457  
  3. Learn More about Stoichiometry brainly.com/question/9743981

Keywords: stoichiometry, theoretical yield, actual yield

3 0
2 years ago
Find the concentraion of [OH-] in a .003 M solution of HBr
nasty-shy [4]

ASDKCODWLQMXNKWAKWKQKSMXSJ

5 0
2 years ago
Calculate the ph of a 0.40 m solution of sodium benzoate (nac6h5coo) given that the ka of benzoic acid (c6h5cooh) is 6.50 x 10-5
asambeis [7]
Hello!

The dissociation reaction for Benzoic Acid is the following:

C₆H₅COOH + H₂O ⇄ C₆H₅COO⁻ + H₃O⁺

The Ka expression is the following and we clear for the concentration of H₃O⁺(X) assuming that the dissociation is little so we can rule it out in the denominator of the equation:

Ka= \frac{[C_6H_5COO^{-}]*[H_3O^{+}] }{[C_6H_5COOH]}=\frac{X*X }{0,40 -X}  (assume: 0,40-X\approx0,40)\\  \\ X= \sqrt{0,40*6,50*10^{-5} }=0.00510M \\  \\ pH=-log([H_3O^{+}]=2,29

So, the pH of this Benzoic Acid solution is 2,29

Have a nice day!


3 0
3 years ago
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