3 years ago

Which activity might help to increase the validity of this experiment?

1 answer:

5 0

A should be the answer because the more you test an experiment the more data you have to rely on changing the experiment would cause you to have different outcomes making the results different and unreliable so B, C, and D is not going to be the answer Hope this helps

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Question 34 (1 point)

Grace [21]

Answer:

8.33 atm

Explanation:

Xe is 5 out of (4+5) or 5 / 9 ths of the gas present

5/9 * 15 atm = 8.33 atm

4 0

1 year ago

If alcohol is ingested rapidly, the _____ will rise sharply within twenty minutes. blood pressure ABC LD50 BAC

GaryK [48]

The Blood Alcohol Level (BAC) will rise. Because it usually peaks about 30 minutes after drinking, a sharp spike at 20 minutes makes sense.

BAC.

5 0

3 years ago

If 15 g of C₂H₆ reacts with 60.0 g of O₂, how many moles of water (H₂O) will be produced?

IceJOKER [234]

Answer:

$n_{H_2O}=1.5molH_2O$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O$

Next, we identify the limiting reactant by computing the available moles of ethane and the moles of ethane consumed by 60.0 grams of oxygen:

$n_{C_2H_6}^{available}=15g*\frac{1mol}{30g} =0.50molC_2H_6\\n_{C_2H_6}^{reacted}=60.0gO_2*\frac{1molO_2}{32gO_2}*\frac{2molC_2H_6}{7molO_2} =0.536molC_2H_6$

Thus, we notice there are less available moles, for that reason, the ethane is the limiting reactant. Finally, we can compute the produced moles of water by:

$n_{H_2O}=0.50molC_2H_6*\frac{6molH_2O}{2molC_2H_6}\\\\n_{H_2O}=1.5molH_2O$