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2 years ago

1. How many grams of glucose are needed to prepare 400mL of 5% glucose solution?

Chemistry

1 answer:

tankabanditka [31]2 years ago

4 0

Answer:

A. 5g

Explanation:

Hopefully this helps

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What mass of ammonia can be produced if 13.4 grams of nitrogen gas reacted ?

aivan3 [116]

Answer:

If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia

Explanation:

Step 1: Data given

Mass of nitrogen gas (N2) = 13.4 grams

Molar mass of N2 = 28 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: Calculate moles of N2

Moles N2 = Mass N2 / molar mass N2

Moles N2 = 13.4 grams / 28.00 g/mol

Moles N2 = 0.479 moles

Step 4: Calculate moles of NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

For 0.479 moles N2 we'll produce 2*0.479 = 0.958 moles

Step 5: Calculate mass of NH3

Mass of NH3 = moles NH3 * molar mass NH3

Mass NH3 = 0.958 moles * 17.03 g/mol

Mass NH3 = 16.3 grams

If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia

3 0

3 years ago

A or B or C or D, fassst plzzz

olga nikolaevna [1]

Answer:

a.option is the correct answer

4 0

2 years ago

Read 2 more answers

I really need help it's urgent help ! Don't ignore it .Your work will be appreciated and don't spam

gtnhenbr [62]

ツ here your answer

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A)Potassium bromide(aq) + Barium iodide(aq) → Potassium iodide(aq) + Barium bromide(s)

2KBr(aq)+BaI2(aq) → 2KI(aq)+BaBr2(s)

B)Balance the Chemical Equation for the reaction of calcium carbonate with hydrochloric acid:
CaCO3+ HCl -> CaCl2 + CO2 + H2O To balance chemical equations we need to look at each element individually on both sides of the equation. calcium carbonate is a chemical compound with the formula CaCO3.

Mark me in brainlist

8 0

2 years ago

The reaction of 5.40 g of carbon with excess O2 yields 13.6 g of CO2. What is the percent yield of this reaction?

vazorg [7]

C + O2= CO2
$n = \frac{m}{mw}$
$n = \frac{5.4}{12} \\ n = 0.45 \: mol \: of \: carbon$
$n = \frac{13.6}{12 + 16 \times 2} \\ n = \frac{13.6}{44} \\ n = 0.31 \: mol \: of \: carbon \: dioxide$
CO2 is limit
5.4-3.72= 1.68 g of C is excess
5.4 g = 100%
3.72 g = x
x=68.9 %

4 0

3 years ago

-Before we begin each flame test we will

ludmilkaskok [199]

Answer:Poop

Explanation:poop

7 0

3 years ago