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Archy [21]
2 years ago
10

What is true of both galvanic and electrolytic cells?

Chemistry
2 answers:
omeli [17]2 years ago
8 0

Answer:

reduction happens at the cathode

Explanation:

Just took the test.

Dmitrij [34]2 years ago
5 0

Answer:

Basically A.

Explanation:

A p E x

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6. Would you describe each of these temperatures as warm, hot, or cold?
mash [69]

Answer:

b: Hot

a: Cold

c: Cold

d: Warm

e: Warm

f: Cold

g: Hot

Explanation:

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2 years ago
For each compound determine whether it is organic or inorganic. C10h16kno9s2 naaso2 hsicl3 (ch)4as2 (bio)2co3 h2p2o7 h2o co2 c6h
Sindrei [870]

a. Organic: C₁₀H₁₆KNO₉S₂; (CH₃)₄As₂; C₆H₁₂O₆

b. Inorganic: NaAsO₂; HSiCl₃; (BiO)₂CO₃; H₂P₂O₇; H₂O; CO₂

Compounds containing <em>both C and H</em> are organic.

Compounds that are <em>not organic</em> are inorganic.

8 0
3 years ago
Displaement of ammonia from its salt
Dvinal [7]
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3 years ago
The combustion of acetylene gas is represented by this equation: 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
MAVERICK [17]

Answer:

Approximately 2.46\; \rm mol.

Explanation:

Make use of the molar mass data (M({\rm C_2H_2}) = 26.04\; \rm g \cdot mol^{-1}) to calculate the number of moles of molecules in that 64.0\; \rm g of \rm C_2H_2:

\begin{aligned}n({\rm C_2H_2}) &= \frac{m({\rm C_2H_2})}{M} \\ &= \frac{64.0\; \rm g}{26.04\; \rm g\cdot mol^{-1}}\approx 2.46\; \rm mol\end{aligned}.

Make sure that the equation for this reaction is balanced.

Coefficient of \rm C_2H_2 in this equation: 2.

Coefficient of \rm H_2O in this equation: 2.

In other words, for every two moles of \rm C_2H_2 that this reaction consumes, two moles of \rm H_2O would be produced.

Equivalently, for every mole of \rm C_2H_2 that this reaction consumes, one mole of \rm H_2O would be produced.

Hence the ratio: \displaystyle \frac{n({\rm H_2O})}{n({\rm C_2H_2})} = \frac{2}{2} = 1.

Apply this ratio to find the number of moles of \rm H_2O that this reaction would have produced:

\begin{aligned}n({\rm H_2O}) &= n({\rm C_2H_2}) \cdot \frac{n({\rm H_2O})}{n({\rm C_2H_2})} \\ &\approx 2.46\; \rm mol \times 1 = 2.46\; \rm mol\end{aligned}.

3 0
3 years ago
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