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3 years ago

PLEASE SHOW WORK and inlcude units on each number

Chemistry

1 answer:

Sedbober [7]3 years ago

4 0

The limiting reagent is H₂SO₄

calculation

Step 1 :write the equation for reaction

2 NaOH + H₂SO₄ → Na₂SO₄ + 2 H₂O

Step 2: use the mole ratio to determine the moles of product produced from each reactant

that is from equation above,

NaOH : Na₂SO₄ is 2 :1 therefore the moles of Na₂So₄

= 10.0 moles x 1/2 = 5.0 moles

H₂SO₄ :Na₂SO₄ is 1:1 therefore the moles of Na₂SO₄ is also = 3.50 moles

H₂SO₄ is the limiting reagent since it produces less amount of Na₂SO₄

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You have 55 grams of glucose(C6H12O6). How many grams of oxygen do you have?

zubka84 [21]

Answer:

Mass of oxygen in glucose = 29.3g

Explanation:

Mass of glucose given is 55grams.

We are to find the mass of oxygen in this compound.

In the compound we have 6 atoms of oxygen.

Solution

To find the mass of oxygen in glucose, we calculate the formula mass of glucose. We now divide the formula mass of the oxygen atom with that of the glucose and multiply by the given mass to find the unkown mass.

Atomic mass of C = 12g

H = 1g

O = 16g

Formula mass of C₆H₁₂O₆ = {(12x6) + (1x12) + (16x6)} = 180

Mass of O in glucose = $\frac{6x16}{180}$ x 55

= $\frac{96}{180}$ x 55

= 0.53 x 55

Mass of oxygen in glucose = 29.3g

8 0

3 years ago

A sample of a pure compound that weighs 60.3 g contains 20.7 g Sb (antimony) and 39.6 g F (fluorine). What is the percent compos

Volgvan

Answer:

The percent composition of fluorine is 65.67%

Explanation:

Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.

That is, the percentage composition is the percentage by mass of each of the elements present in a compound.

The calculation of the percentage composition of an element is made by:

$percent composition element A=\frac{total mass of element A}{mass of compound} *100$

In this case, the percent composition of fluorine is:

$percent composition of fluorine=\frac{39.6 g}{60.3 g} *100$

percent composition of fluorine= 65.67%

The percent composition of fluorine is 65.67%

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4 years ago

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allsm [11]

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3 years ago

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Monica [59]

50% - you have two bb pairs

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umka2103 [35]

Answer:

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