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ololo11 [35]
3 years ago
13

ANYONE PLEASE HELP ME WITH MY CHEMISTRY HOMEWORK I REALLY NEED THE ANSWER RIGHT NOW I HOPE Y’ALL CAN HELP ME:(

Chemistry
1 answer:
nevsk [136]3 years ago
7 0

Answer:

what grade is this ?

Explanation:

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It was a power supply of a circuit that was horizontal of the ladder where they controlled the circuit.
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What was madame curry famous for
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For the discovery of the elements polonium and radium. Prohibited from higher education in her native Poland (then controlled by Russia), she moved to Paris in 1891 and studied at the Sorbonne.
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How many grams of H2O will be formed when 32.0 g H2 is mixed with 84.0 g of O2 and allowed to react to form water
Zarrin [17]

Answer:

94.58 g of H_2O

Explanation:

For this question we have to start with the reaction:

H_2~+~O_2~->~H_2O

Now, we can balance the reaction:

2H_2~+~O_2~->~2H_2O

We have the amount of H_2  and the amount of O_2 . Therefore we have to find the limiting reactive, for this, we have to follow a few steps.

1) Find the moles of each reactive, using the molar mass of each compound (H_2~=~2~g/mol~~O_2=~32~g/mol ).

2) Divide by the coefficient of each compound in the balanced reaction ("2" for H_2 and "1" for O_2).

<u>Find the moles of each reactive</u>

32.0~g~H_2\frac{1~mol~H_2}{2~g~of~H_2}=15.87~mol~H_2

84.0~g~of~O_2\frac{1~mol~of~O_2}{32~g~of~O_2}=2.62~mol~of~O_2

<u>Divide by the coefficient</u>

<u />

\frac{15.87~mol~H_2}{2}=7.94

\frac{2.62~mol~of~O_2}{1}=2.62

The smallest values are for H_2, so hydrogen is the limiting reagent. Now, we can do the calculation for the amount of water:

32.0~g~H_2\frac{1~mol~H_2}{2~g~of~H_2}\frac{2~mol~H_2O}{2~mol~H_2}\frac{18~g~H_2O}{1~mol~H_2O}=94.58~g~H_2O

We have to remember that the molar ratio between H_2O and H_2 is 2:2 and the molar mass of H_2O is 18 g/mol.

6 0
3 years ago
A) melissa planted more seeds than juan.
kolbaska11 [484]
C the two students watered the seeds differently
3 0
3 years ago
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Calcium carbide (CaC2) reacts with water to form acetylene (C2H2) and Ca(OH)2.
Sloan [31]
The answer is -60.57. With appropriate sig figs it becomes -60.6 KJ. 

<span>I think you were on the right track. However you need to multiply H2O by its coefficient from the balanced equation. Then you also need to keep in mind the signs. </span>

<span>Maybe it's easier if I write what I did out: </span>

<span>2H2O > 2H2 +O2...............-H= 2* -285.83 </span>
<span>Ca + O2 + H2 > Ca(OH)2....H= -986.2 </span>
<span>2C + H2 > C2H2.................H= 226.77 </span>

<span>The above H's stand for standard enthalpy of formations. These can be found in textbook appendix. Notice the negative infront of the enthalpy (H) for H2O. This is to remind me/you that the heat lost is gained in the rxn. </span>

<span>So then you add them up. 226.77 - 986.2 + (2*285.83) = -187.77 </span>

<span>Add back the total enthalpy that is given in the question -187.77+127.2 = -60.57 </span>

<span>If signs cross you up find a way to remember that works for you. </span>
4 0
4 years ago
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