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Hatshy [7]
2 years ago
6

The ____ is found on the right side of the arrow in a chemical reaction

Chemistry
1 answer:
s344n2d4d5 [400]2 years ago
4 0
The PRODUCT is found on the right side of the arrow in a chemical reaction.
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Balance the following: ___ AlBr3+ ___ K---> ___KBr+ ___ Al
Nataliya [291]
___AlBr3 + ___K -> ___KBr + ___ Al

1 AlBr3 + 3K -> 3KBr + 1 Al

hope this helps............
6 0
3 years ago
What do you think the density of a person might be? Explain
Verizon [17]
So this can explain the density of a person very clearly.

7 0
3 years ago
Which of the following is the balanced equation for this process?
Hunter-Best [27]
The answer is A.Hope this helps
8 0
3 years ago
Part A
den301095 [7]

Answer:

n_{Cl_2}=0.3molCl_2

Explanation:

Hello there!

In this case, according to the given chemical reaction whereas the sodium chloride is in a 2:1 mole ratio with chlorine, the required moles of the later are computed as shown below:

n_{Cl_2}=0.6molNaCl*\frac{1molCl_2}{2molNaCl}

So we cancel out the moles of NaCl to obtain:

n_{Cl_2}=0.3molCl_2

Best regards!

3 0
2 years ago
Information-
tekilochka [14]

Answer:

So first thing to do in these types of problems is write out your chemical reaction and balance it:

Mg + O2 --> MgO

Then you need to start thinking about moles of Magnesium for moles of Magnesium Oxide. Based on the above equation 1 mole of Magnesium is needed to make one mole of Magnesium Oxide.

To get moles of magnesium you need to take the grams you started with (.418) and convert to moles by dividing by molecular weight of Mg (24.305), this gives you .0172 moles of Mg.

The theoretical yield would be the assumption that 100% of the magnesium will be converted into Magnesium Oxide, so you would get, based on the first equation, .0172 mol of MgO. Multiplying this by the molecular weight of MgO (24.305+16) gives us .693 g of MgO.

The percent yield is what you actually got in the experiment, and for this you subtract off the total mass from the crucible mass, or 27.374 - 26.687, which gives .66 g of MgO obtained.

Percent yield is acutal/theoretical, .66/.693, or 95.24%.

I'll let you do the same for the second trial, and average percent yield is just an average of the two trials percent yield.

Hope this helps.

6 0
3 years ago
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