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kirill115 [55]
3 years ago
10

If 10 grams of water are added to 5 grams of salt, how much salt water will be made?

Chemistry
2 answers:
KiRa [710]3 years ago
8 0

Answer : The amount of salt water made will be, 15 grams.

Explanation : Given,

Mass of water = 10 grams

Mass of salt = 5 grams

Now we have to determine the amount of salt water.

Mass of salt water = Mass of water + Mass of salt

Mass of salt water = 10 g + 5 g

Mass of salt water = 15 g

Therefore, the amount of salt water made will be, 15 grams.

charle [14.2K]3 years ago
7 0
<span>the mass should increase by exactly the 5g you added. </span>
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A. Which reactant is the limiting reagent?
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b. m_{ZnBr_2}^{by\ Zn}=162.61gZnBr_2

c. m_{Br_2}^{leftover}=6.6g

Explanation:

Hello there!

a. In this case, when zinc metal reacts with bromine, the following chemical reaction takes place:

Zn+Br_2\rightarrow ZnBr_2

Thus, since zinc and bromine react in a 1:1 mole ratio, we can compute their reacting moles to identify the limiting reactant:

n_{Zn}=47.2g*\frac{1mol}{65.38g} =0.722molZn\\\\n_{Br_2}=122g*\frac{1mol}{159.8g} =0.763molBr_2

Thus, since zinc has the fewest moles we infer it is the limiting reactant.

b. Here, we compute the grams of zinc bromide via both reactants:

m_{ZnBr_2}^{by\ Zn}=0.722molZn*\frac{1molZnBr_3}{1molZn} *\frac{225.22gZnBr_2}{1molZnBr_2} =162.61gZnBr_2\\\\m_{ZnBr_2}^{by\ Br_2}=0.763molBr_2*\frac{1molZnBr_3}{1molBr_2} *\frac{225.22gZnBr_2}{1molZnBr_2} =171.95gZnBr_2

That is why zinc is the limiting reactant, as it yields the fewest moles of zinc bromide product.

c. Here, since just 0.722 mol of bromine would react, we compute the corresponding mass:

m_{Br_2}^{reacted}=0.722molBr_2*\frac{159.8gBr_2}{1molBr_2} =115.4gBr_2

Thus, the leftover of bromine is:

m_{Br_2}^{leftover}=122g-115.4g\\\\m_{Br_2}^{leftover}=6.6g

Best regards!

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