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2 years ago

Empirical formula for C2H4O2

Chemistry

1 answer:

bezimeni [28]2 years ago

6 0

Answer:

Explanation:

The molecular formula of CH 3 COOH can be written as C2H4O2 . The ratios of C: H:O. are 2:4:2. The highest common factor of 2 and 4 is 2.

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A 2.600×10−2 M solution of glycerol (C3H8O3) in water is at 20.0∘C. The sample was created by dissolving a sample of C3H8O3 in w

sesenic [268]

Answer:

A. 0.2395 w/w %

B. 2394ppm

Explanation:

A. To find concentrationin percent by mass of the solution we need to calculate mass of glycerol and mass of water. The formula is:

Mass glycerol / Total mass * 100

<em>Mass glycerol:</em>

The solution is 2.6x10⁻²moles / L. As there is 1L of solution there are 2.6x10⁻² moles of glycerol. In mass (Using molar mass glycerol: 92.09g/mol):

2.6x10⁻² moles of glycerol * (92.09g / mol) = 2.394g glycerol

<em>Mass of water:</em>

998.9mL and density = 0.9982g/mL:

998.9mL * (0.9982g/mL) = 997.1g of water.

That means percent by mass is:

% by mass: 2.394g / (997.1g + 2.394g) * 100 = 0.2395 w/w %

B. Parts per million are mg of glycerol per L of solution. As in 1L there are 2.394g. In mg:

2.394g * (1000mg / 1g) = 2394mg:

Parts per million: 2394mg / L = 2394ppm

3 0

3 years ago

In the equation below, which of the following is true?<br><br> 2Fe + 6HBr yields 2FeBr3 + 3H2

eimsori [14]

2Fe+ 6HBr ---> 2FeBr3 +3H2
Fe is a reducing agent is what that is true in equation above This is because it oxidation state move from zero to positive three.Reducing agent is an element or compound that loses or donate an electron to another chemical species in a redox chemical reaction. Fe is oxidized since it loses electron in the redox reaction

8 0

3 years ago

Calculate the percent activity of the radioactive isotope iodine-131 remaining after 3 half-lives.

vovangra [49]

Half life of a radio-active material is refereed to the time required for reducing its activity to 50 %.

Now, when I-131 passes 1st half life, it's activity will reduce to half i.e. 50%

When, I-131 passes 2nd half life, activity of I-131 will be reduced to 25%

On passing, 3rd half-life, percent activity of the I-131 <span>remaining will be 12.5 %.</span>

3 0

4 years ago

2. What is the mass of one mole of Na?

borishaifa [10]

Sodium, Atomic mass: 22.989769 g
You can see in a periodic table

5 0

3 years ago

Pure iron, Fe , can be produced from an ore called hematite, Fe2O3 , by reaction with carbon at high temperatures. 2Fe2O3+3C⟶4Fe

Sophie [7]

70 tons of hematite (Fe₂O₃) is needed to produce 49 tons of iron (Fe).

We'll begin by calculating the tons of hematite (Fe₂O₃) that reacted and the tons of Fe produced from the balanced equation. This is illustrated below:

2Fe₂O₃ + 3C —> 4Fe + 3CO₂

Molar mass of Fe₂O₃ = (56×2) + (16×3) = 160 g/mol

Mass of Fe₂O₃ from the balanced equation = 2 × 160 = 320 g

Divide by 907185 to express in ton

320 / 907185 = 0.000353 ton

Molar mass of Fe = 56 g/mol

Mass of Fe from the balanced equation = 56 × 4 = 224 g

Divide by 907185 to express in ton

224 / 907185 = 0.000247 ton

<h3>SUMMARY </h3>

From the balanced equation above,

0.000353 ton of Fe₂O₃ reacted to produce 0.000247 ton of Fe.

Finally, we shall determine the tons of hematite (Fe₂O₃) needed to produce 49 tons of Fe. This can be obtained as follow:

From the balanced equation above,

0.000353 ton of Fe₂O₃ reacted to produce 0.000247 ton of Fe.

Therefore,

X ton of Fe₂O₃ will react to produce 49 tons of Fe i.e

X ton of Fe₂O₃ = $\frac{0.000353 * 49}{0.000247} \\\\$

X ton of Fe₂O₃ = 70 tons

Thus, 70 tons of hematite (Fe₂O₃) is needed to produce 49 tons of Fe.

Learn more: brainly.com/question/15343472

5 0

3 years ago