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Natasha_Volkova [10]
3 years ago
14

ANSWER FAST PLZ 25 POINTS!!!!!!!!!!!!!!!

Chemistry
1 answer:
Scilla [17]3 years ago
8 0

Answer:

i think the answer is C)

Explanation:

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Determine the empirical formula for a compound that contains 15.8% carbon and 84.2% sulfer
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Carbon Disulfide CS2

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How many grams of glucose are needed to prepare 144.3 mL of a 1.4%(m/v) glucose solution?
Montano1993 [528]

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2.0202 grams

Explanation:

1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.

so ?g glucose = 144.3 mL soln

Now apply the conversion factor, and you have:

?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).

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William measures a test tube and finds that the mass of the test tube is 5g. He places the lone reactant in the test tube and fi
iogann1982 [59]

86 percent is the percent yield for this experiment if he expected to produce 5g of product.

Explanation:

Given that:

mass of test tube = 5 grams

mass of test tube + reactant is 12.5 grams

mass of reactant = ( mass of test tube + reactant ) - (mass of test tube)

mass of reactant = 12.5 -5

                             = 7.5 grams

when 7.5 grams of reactant is heated mass of test tube was found to be 9.3 grams.

so mass of product formed = 9.3 - 5

                                             = 4. 3 grams of product is formed (actual yield)

However, he expected the product to be 5 grams (theoretical yield)

Percent yield = \frac{actual yield}{theoretical yield} x 100

          putting the values in the formula:

percent yield = \frac{4.3}{5} x 100

                     = 86 %

86 percent is the percent yield.

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3 years ago
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