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sukhopar [10]
3 years ago
12

John Dalton believed which of the following about atoms?

Chemistry
1 answer:
8090 [49]3 years ago
3 0
Atoms are real even though they're invisible.
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6. How will you obtain ? (a) Magnesium oxide from magnesium. (b) Silver chloride from silver nitrate. (c) Nitrogen dioxide from
Reil [10]

Answer:

a) reaction with oxygen

2mg +o2---------2mgo

b) Agno3+NaCl ----------AgCl+NaNo3

8 0
2 years ago
The bonds in the compound MgSO4 can be described as
Butoxors [25]
C. Sulfur and oxygen (non metals) forms a covalent bond while the magnesium (a metal) will react with both non metals to form an ionic bond
7 0
4 years ago
True or False. Solutions for which water is the solvent are called aqueous solutions.
Stolb23 [73]
This is true. Water is the solvent in aqueous solutions
5 0
3 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.

7 0
3 years ago
A solution is made by mixing 33.0 ml of ethanol, C2H6O and 67.0 ml of water. Assuming ideal behavior, what is the vapor pressure
ivann1987 [24]
Grams ethanol = 33 ml times .789 gms/ml = 26.037 gms 

<span>Moles ethanol = 26.037 gms / 46 gms/mole = .57 moles </span>

<span>Moles water = 67 ml or 67 grams/18 gms/mole = 3.22 moles </span>

<span>total moles = .57 + 3.72 = 4.29 moles </span>

<span>Mole fraction ethanol = .57 moles ethanol / 4.29 moles total = 0.13</span>

<span>Moles fraction water = 3.72 moles water / 4.29 moles total = 0.87</span>

<span>Partial pressure of ethanol = mole fraction ethanol (.13) _ times VP ethanol 43.9 torr) = 5.707 torr </span>

<span>partial pressure water = mole fraction water .87) times VP water (l7.5 torr) = 15.23 torr </span>

<span>Total vapor pressure over solution = 5.71 torr + 15.23 torr = 20.94 torr</span>
7 0
3 years ago
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