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krok68 [10]
3 years ago
11

Which would aid a student in preparedness for a lab?

Chemistry
2 answers:
malfutka [58]3 years ago
7 0
I would say C. Telling the teacher about a chemical spill
mote1985 [20]3 years ago
4 0

Answer: cleaning up after the lab, D.

Explanation:

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9 is clay, silt, sand in that order

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KBr + AgNO3 → AgBr + KNO3<br><br><br> Which two chemicals are products in the chemical reaction?
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Products are AgBr and KNO3

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Name the parts of an electromagnet
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The iron core, copper wire, and an electricity source.

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Explain how human activity can affect the carbon cycle.
Eva8 [605]

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              Human activities badly affect carbon cycle. Activities such as burning fossil fuel and deforestation have begun to effect on carbon cycle and the rise of carbon dioxide in atmosphere.

Explanation:

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4 0
3 years ago
Read 2 more answers
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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