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krok68 [10]
3 years ago
15

What is the percent of Cu in CuSO4?

Chemistry
1 answer:
hoa [83]3 years ago
7 0
Anhydrous Copper sulfate is 39.81 percent copper and 60.19 percent sulfate by mass, and in its blue, hydrous form, it is 25.47% copper, 38.47% sulfate (12.82% sulfur) and 36.06% water by mass.
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Breathe in air container more ____ than breathed out air
TEA [102]

Answer:

c

Explanation:

carbon dioxide ......

5 0
3 years ago
What is migrate? Help me pls
Juliette [100K]
Means moving from one place to the other
8 0
3 years ago
Read 2 more answers
In a single displacement reaction between sodium phosphate and barium, how much of each product (in grams) will be formed from 1
weeeeeb [17]

Answer:

A. 3.36g of Na.

B. 14.62g of Ba3(PO4)2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

3Ba + 2Na3PO4 → 6Na + Ba3(PO4)2

Next, we shall determine the mass of Ba that reacted and the mass of Na and Ba3(PO4)2 produced from the equation.

This is illustrated below:

Molar Mass of Ba = 137g/mol

Mass of Ba from the balanced equation = 3 x 137 = 411g

Molar mass of Na = 23g/mol

Mass of Na from the balanced equation = 6 x 23 = 138g

Molar mass of Ba3(PO4)2 = (3 x 137) + 2[31 + (4x16)] = 411 + 2[31 + 64] = 601g/mol

Mass of Ba3(PO4)2 from the balanced equation = 1 x 601 = 601g

Summary:

From the balanced equation above,

411g of Ba reacted to produce 138g of Na and 601g of Ba3(PO4)2.

A. Determination of the mass of Na produced by reacting 10g of Ba.

From the balanced equation above,

411g of Ba reacted to produce 138g of Na.

Therefore, 10g of Ba will react to produce = (10 x 138)/411 = 3.36g of Na.

Therefore, 3.36g of Na is produced.

B. Determination of the mass of Ba3(PO4)2 produced by reacting 10g of Ba.

From the balanced equation above,

411g of Ba reacted to produce 601g of Ba3(PO4)2.

Therefore, 10g of Ba will react to produce = (10 x 601)/411 = 14.62g of Ba3(PO4)2.

Therefore, 14.62g of Ba3(PO4)2 is produced.

7 0
3 years ago
An aqueous basic solution has a concentration of 0. 050 m and kb is 4. 4 × 10^-4. What is the concentration of hydronium ion in
Alex_Xolod [135]

An aqueous basic solution has a concentration of 0. 050 m and kb is 4. 4 × 10^-4, the concentration of hydronium ion in this solution (m) is 2.234 × 10⁻¹² M.

Methylamine is an amine which is an organic weak base. Its chemical formula is CH₃NH₂. When it undergoes hydrolysis wherein water is acting as an acid, the reaction would be

CH₃NH₂ + H₂O ⇆  CH₃NH₃ + OH⁻

Then, we use the ICE analysis which stands for Initial-Change-Equilibrium.

CH₃NH₂ + H₂O ⇆  CH₃NH₃ + OH⁻

Initial                    0.05          -                 0         0

Change                 -x                              +x       +x

----------------------------------------------------------------------------

Equilibrium         0.05-x                           x          x

Then, we use the equation for the equilibrium constant of basicity.

Kb = [CH₃NH₃][OH⁻]/[CH₃NH₂] = 4.4×10⁻⁴

4.4×10⁻⁴ = [x][x]/[0.05-x]

[x] = 4.4756×10⁻³

Here, variable x denotes the number of moles of the substance which is involved in the reaction. Since the equilibrium amount of OH⁻ is equal to x, then the concentration of OH⁻ is also 4.4756×10⁻³. Thus,

pOH = -log[OH⁻]

pOH = -log[4.4756×10⁻³] = 2.35

The relationship between pOH and pH is that pH + pOH = 14. Thus,

pH = 14 - 2.35 = 11.65

pH = -log[H⁺]

11.65 = -log[H⁺]

[H⁺] = 2.234 × 10⁻¹² M

Thus, we find the concentration of solution is 2.234 × 10⁻¹² M.

Learn more about aqueous solution: brainly.com/question/11097800

#SPJ4

5 0
2 years ago
Consider the following reaction at 298K.
lakkis [162]

The true statements are;

  • K < 0
  • Eocel  < 0

<h3>What is a redox reaction?</h3>

We define a redox reaction as one in which a specie is oxidized and another is reduced.

Now;

Eo cell = cell potential = -0.13 V - (+0.34 V) = -0.47 V

n =number of moles of electrons = 2 mole of electrons

K = equilibrium constant

ΔG = change in free energy

Eo cell = 0.0592/n log K

-0.47 =  0.0592/2 log K

log K =  -0.47  * 2/0.0592

K = 1.3 * 10^-16

ΔG = -nFEo cell

ΔG = -(2 * 96500 * -0.47)

ΔG = 90.7kJ

Learn more about Ecell:brainly.com/question/10203847

#SPJ1

8 0
2 years ago
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