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3 years ago

I BEG I NEED HELP

Chemistry

2 answers:

NNADVOKAT [17]3 years ago

6 0

Answer:

The percent yield of the reaction is 82.5%

Explanation:

Let's work with moles to get the percent yield.

Mass / Molar mass =

652.5 g / 158.03 g/m = 4.13 moles

If the theoretical yield of the reaction is 5 moles but we only made 4.13 moles, the percent yield will be:

(Produced yield / Theoretical yield) . 100 =

(4.13 / 5) . 100 = 82.5 %

Whitepunk [10]3 years ago

3 0

Answer:

The % yield is 82.58 %

Explanation:

Step 1: Data given

theoretical yield = 5.00 moles of KMnO4

Molar mass of KMnO4 = 158.03 g/mol

Mass of KMnO4 produced = 652.5 grams

Step 2: Calculate mass of 5.00 moles KMnO4

Mass = moles * molar mass

Mass KMnO4 = 5.00 moles * 158.03 g/mol

Mass KMnO4= 790.15 grams = theoretical yield

Step 3: Calculate % yield

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

% yield = (652.5 / 790.15)* 100 %

% yield = 82.58%

The % yield is 82.58 %

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A chemist titrates 110.0 mL of a 0.7684 M methylamine (CH3NH2) sotion with 0.4469 M HNO3 solution at 25 °C. Calculate the pH at

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$CH_{3}NH_{3}^{+} + H_{2}O \rightleftharpoons CH_{3}NH_{2} + H_{3}O^{+}$

As, $k_{a} = \frac{k_{w}}{k_{b}}$

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