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nika2105 [10]
3 years ago
11

martha needs to determine the density of gasoline she has determined the mass to be 1250g and the volume to be 1500ml what is th

e density of the gasoline
Chemistry
1 answer:
qaws [65]3 years ago
7 0

0.833 g/mL

Density = mass/volume = (1250 g)/(1500 mL) = 0.833 g/mL

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How many ions are in 1.50 grams of Na+?
Jobisdone [24]

Answer:

= 0.0652 moles

Explanation:

I hope it helps you

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2 years ago
When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
Debora [2.8K]

Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

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3 years ago
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Answer and Explanation:

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3 years ago
What is an atomic number?
padilas [110]
An atomic number is <span>the number of protons in the nucleus of an atom, which determines the chemical properties of an element and its place in the periodic table or chart.</span>
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3 years ago
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