No question................................

A chemical company makes two brands of antifreeze. the first brand is 70% pure antifreeze, and the second brand is 95% pure anti

madreJ [45]

Answer is: 56 gallons of 70% antifreeze and 84 gallons of 95% antifreeze.

ω₁ = 70% ÷ 100% = 0.7; 70% pure antifreeze.

ω₂ = 95% ÷ 100% = 0.95.

ω₃ = 85% ÷ 100% = 0.85.
V₁ = ?; volume of 70% antifreeze.

V₂ = ?; volume of 95% antifreeze.
V₃ = V₁ + V₂.
V₃ = 140 gal.

V₁ = 140 gal - V₂.
ω₁ · V₁ + ω₂ ·V₂ = ω₃ · V₃.

0.70 · (140 gal - V₂) + 0.95 · V₂ = 0.85 · 140 gal.

98 gal - 0.7V₂ + 0.95V₂ = 119 gal.

0.25V₂ = 21 gal.

V₂ = 21 gal ÷ 0.25.

V₂ = 84 gal.

V₁ = 140 gal - 84 gal.

V₁ = 56 gal.

4 0

3 years ago

What is Zeff of phosphorus

professor190 [17]

Answer:

Explanation:

Here is an illustration showing how to "calculate" +5 as the effective nuclear charge (Zeff) for phosphorus. We are not implying that the electrons are in orbits here, this is simply showing inner shell electrons that shield the outer valence electrons from the full nuclear charge.

5 0

2 years ago

Draw a Lewis structure for the following molecules on scratch paper. Then give the number of bonds between the C and N in each.

Yuliya22 [10]

Refer to the attachments

6 0

3 years ago

9.50 L of Fluorine gas at STP reacts with 6.945x1024 formula units of Chromium (VI) chloride.

lorasvet [3.4K]

Answer:

Again please specify your question, you still haven't indicated what you want to be solved.

by formula units I assume you mean atoms?

I'll edit my answer accordingly if you could specify down in the comments.

8 0

2 years ago

According to the experimental procedure of Experiment F1, 135 microliters of acetophenone (120.15 g/mol, 1.03 g/mL) was reacted

Delicious77 [7]

Answer: The theoretical yield of 4-nitrochalcone is, $2.13\times 10^2$

Explanation : Given,

Volume of acetophenone = 135 microliters = 135 × 10⁻⁶ L = 0.135 mL

conversion used : (1 microliter = 10⁻⁶ L) and (1 L = 1000 mL)

Density of acetophenone = 1.03 g/mL

Mass of acetophenone = Density × Volume = 1.03 g/mL × 0.135 mL = 0.139 g

Mass of 4-nitrobenzaldehyde = 127 mg = 0.127 g

Conversion used : (1 mg = 0.001 g)

First we have to calculate the moles of acetophenone and 4-nitrobenzaldehyde

$\text{Moles of acetophenone}=\frac{\text{Given mass acetophenone}}{\text{Molar mass acetophenone}}$

$\text{Moles of acetophenone}=\frac{0.139g}{120.15g/mol}=0.00116mol$

and,

$\text{Moles of 4-nitrobenzaldehyde}=\frac{\text{Given mass 4-nitrobenzaldehyde}}{\text{Molar mass 4-nitrobenzaldehyde}}$

$\text{Moles of 4-nitrobenzaldehyde}=\frac{0.127g}{151.12g/mol}=0.000840mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$C_8H_8O+C_7H_5NO_3\rightarrow C_{15}H_{11}NO_3$

From the balanced reaction we conclude that

As, 1 mole of 4-nitrobenzaldehyde react with 1 mole of acetophenone

So, 0.000840 mole of 4-nitrobenzaldehyde react with 0.000840 mole of acetophenone

From this we conclude that, acetophenone is an excess reagent because the given moles are greater than the required moles and 4-nitrobenzaldehyde is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of 4-nitrochalcone

From the reaction, we conclude that

As, 1 mole of 4-nitrobenzaldehyde react to give 1 mole of 4-nitrochalcone

So, 0.000840 mole of 4-nitrobenzaldehyde react to give 0.000840 mole of 4-nitrochalcone

Now we have to calculate the mass of 4-nitrochalcone

$\text{ Mass of 4-nitrochalcone}=\text{ Moles of 4-nitrochalcone}\times \text{ Molar mass of 4-nitrochalcone}$