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sammy [17]
3 years ago
6

The electrolyte in a typical wet storage cell or battery is a mixture of water and A. lead. B. ammonium chloride. C. sulfuric ac

id. D. zinc.
Chemistry
2 answers:
Vilka [71]3 years ago
5 0
The correct answer is C. sulfuric acid.
ludmilkaskok [199]3 years ago
5 0
It's <span>67% water and 33% Sulfuric acid. Therefore the answer is C. Sulfuric Acid </span>
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Which of these choices is a physical property that does not change when the size of the sample changes? mass,volume,density,comb
Shalnov [3]

Answer:

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Explanation:

6 0
3 years ago
What is the number of the group that has the most reactive nonmetals (give number and letter- Ex 1A, 2A, etc.)?​
Kryger [21]

Answer:

Group 7A

Explanation:

The group 7A elements consists of the most reactive non-metals on the periodic table.

This group is known as the group of halogens. They consist of element fluorine, chlorine, bromine, iodine and astatine.

  • The elements in this group have the highest electronegativity values.
  • They have 7 valence electrons and requires just one electron to complete their octets.
  • This way, they are highly reactive in their search for that single electron.
3 0
3 years ago
Hydrogen sulfide burns form sulfur dioxide:
Helga [31]

Answer: 404.04 kJ.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

moles of H_2S

\text{Number of moles}=\frac{26.7g}{34.1g/mol}=0.78moles

2H_2S(g)+3O_2(g)\rightarrow 2SO_2(g)+2H_2O(g)    \Delta H=-1036kJ

According to stoichiometry :

2 moles of H_2S on burning produces = 1036 kJ

Thus 0.78 moles of H_2S on burning produces =\frac{1036kJ}{2}\times 0.78=404.04

Thus the enthalpy change when burning 26.7 g of hydrogen sulfide is 404.04 kJ.

7 0
3 years ago
Commercial manufacturers produce nitric acid (HNO3) by the Ostwald process. The process requires three steps:
Shkiper50 [21]

Answer:

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Explanation:

5 0
3 years ago
CaBr + KOH – Ca(OH), + KBr (balance first) What mass, in grams, of
neonofarm [45]

Answer:

129.73 g of CaBr₂

Explanation:

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

CaBr₂ + 2KOH –> Ca(OH)₂ + 2KBr

Next, we shall determine the mass of CaBr₂ that reacted and the mass of Ca(OH)₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of CaBr₂ = 40 + (80×2)

= 40 + 160

= 200 g/mol

Mass of CaBr₂ from the balanced equation = 1 × 200 = 200 g

Molar mass of Ca(OH)₂ = 40 + 2(16 + 1)

= 40 + 2(17)

= 40 + 34

= 74 g/mol

Mass of Ca(OH)₂ from the balanced equation = 1 × 74 = 74 g

SUMMARY :

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Finally, we shall determine the mass of CaBr₂ that react when 48 g of Ca(OH)₂ were produced. This can be obtained as follow:

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Therefore, Xg of CaBr₂ will react to produce 48 g of Ca(OH)₂ i.e

Xg of CaBr₂ = (200 × 48)/74

Xg of CaBr₂ = 129.73 g

Thus, 129.73 g of CaBr₂ were consumed.

6 0
3 years ago
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