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Tresset [83]
2 years ago
6

Franklin was performing an experiment by combining hydrochloric acid and sodium hydroxide. He measured the mass of his reactant

materials to be 35g. The reaction resulted in the production of salt and water. He reported that his products weighed 32g. Which best describes the results of Franklin’s experiment? No error occurred, some of the products are always lost as heat. An error occurred, the mass of the reactants should equal the mass of the products. An error occurred, the products should weigh more than the reactants. No error occurred, water is not weighed when determining the weight of the products.
Chemistry
2 answers:
ArbitrLikvidat [17]2 years ago
8 0

Answer:

B

Explanation:

VARVARA [1.3K]2 years ago
4 0

Answer:

B

Explanation:

took the test

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iron

Explanation:

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The symbols given to elements are always directly related to their chemical names
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Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide. 4 HCl ( aq ) + MnO
eduard

Answer:

HCl is the limiting reactant.

20.2 grams = theoretical yield Cl2

actual yield = 16.75 grams Cl2

Explanation:

Step 1: Data given

Mass of MnO2 = 43.5 grams

Molar mass MnO2 = 86.94 g/mol

Mass of HCl = 41.5 grams

Molar mass HCl = 36.46 g/mol

Step 2: The balanced equation

4HCl(aq)+MnO2(s)⟶MnCl2(aq)+2H2O(l)+Cl2(g)

Step 3: Calculate moles

Moles = mass / molar mass

Moles MnO2 = 43.5 grams / 86.94 g/mol

Moles MnO2 =  0.500 moles

Moles HCl = 41.5 grams / 36.46 g/mol

Moles HCl = 1.14 moles

Step 4: Calculate the limiting reactant

For 4 moles HCl we need 1 mol MnO2 to produce 1 mol MnCl2, 2 moles H2O and 1 mol Cl2

HCl is the limiting reactant. I will completely be consumed (1.14 moles).

MnO2 is in excess. There will react 1.14 /4 = 0.285 moles

There will remain 0.500 - 0.285 = 0.215 moles MnO2

Step 5: Calculate moles Cl2

For 4 moles HCl we need 1 mol MnO2 to produce 1 mol MnCl2, 2 moles H2O and 1 mol Cl2

For 1.14 moles HCl we'll have 1.14/4 = 0.285 moles Cl2

Step 6: Calculate mass Cl2

Mass Cl2 = moles Cl2 * molar mass Cl2

Mass Cl2 = 0.285 moles * 70.9 g/mol

Mass Cl2 = 20.2 grams = theoretical yield

Step 7: Calculate actual yield

% yield = (actual yield / theoretical yield) *100%

0.829 = actual yield / 20.2 grams

actual yield = 0.829 * 20.2 grams

actual yield = 16.75 grams Cl2

6 0
3 years ago
The name of the ion P3- is phosphoride ion.<br><br> True<br> False
melisa1 [442]
I believe that it would be false. <span>The name of the ion P3- is not phosphoride ion but it is phosphide ion. </span><span>The </span>phosphide ion<span> is P </span>3−<span>, and </span>phosphides<span> of almost every metal in the periodic table are known. They exhibit a wide variety of chemical and physical properties. Hope this answers the question.</span>
5 0
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g Calculate the theoretical yield (in grams) of your product if you start with 0.50 grams of E-stilbene. The molecular weight of
sattari [20]

Answer:

0.9433g

Explanation:

Theoretical yield is defined as the mass produced assuming all reactant reacts producing the product.

Assuming the reaction is 1:1, we need to find the moles of E-stilbene (Reactant). If all reactant reacts, the moles of E-stilbene = Moles of product.

Using the molar mass of the product we can find the theoretical yield as follows:

<em>Moles E-stilbene:</em>

0.50g * (1mol/180.25g) = 0.00277 moles = Moles Product

<em>Mass Product = Theoretical yield:</em>

0.00277 moles * (340.058g/mol) = 0.9433g

4 0
3 years ago
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