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Lana71 [14]
3 years ago
9

There are three isotopes of oxygen o-16 o- 17 and o-18. these neutral atoms all contain?

Chemistry
1 answer:
Jobisdone [24]3 years ago
7 0
The same proton number
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Find the mass of a single one of the molecules or ionic compounds listed below. You must show your work in the correct format an
Lemur [1.5K]
I don’t really know tho
3 0
3 years ago
What mass of CH3COOH is present in a 250 mL cup of 1.25 mol/L solution of vinegar?
aleksley [76]
If    1000 ml (1 L) of CH₃COOH contain 1.25 mol
let  250 ml  of CH₃COOH contain x

⇒  x =  \frac{250 ml * 1.25mol}{1000 ml}
        
        =  0.3125 mol

∴ moles of CH₃COOH in 250ml is 0.3125 mol

Now, Mass = mole  ×  molar mass
        
                   = 0.3125 mol  × [(12 × 2)+(16 × 2)+(1 × 4)] g/mol
 
                   = 18.75 g

∴ Mass of CH₃COOH present in a 250 mL cup of 1.25 mol/L solution of vinegar is <span>18.75 g</span>
4 0
3 years ago
1) Aluminum sulphate can be made by the following reaction: 2AlCl3(aq) + 3H2SO4(aq) Al2(SO4)3(aq) + 6 HCl(aq) It is quite solubl
kolezko [41]

Answer:

88.9%

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AlCl3(aq) + 3H2SO4(aq) —> Al2(SO4)3(aq) + 6HCl(aq)

Step 2:

Determination of the masses of AlCl3 and H2SO4 that reacted and the mass of Al2(SO4)3 produced from the balanced equation.

Molar mass of AlCl3 = 27 + (35.5x3) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar mass of Al2(SO4)3 = (27x2) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4 to produce 342g of Al2(SO4)3.

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4.

Therefore, 25g of AlCl3 will react with = (25 x 294)/267 = 27.53g of H2SO4.

From the calculations made above, we see that only 27.53g out 30g of H2SO4 given were needed to react completely with 25g of AlCl3.

Therefore, AlCl3 is the limiting reactant and H2SO4 is the excess.

Step 4:

Determination of the theoretical yield of Al2(SO4)3.

In this case we shall be using the limiting reactant because it will produce the maximum yield of Al2(SO4)3 since all of it is used up in the reaction.

The limiting reactant is AlCl3 and the theoretical yield of Al2(SO4)3 can be obtained as follow:

From the balanced equation above,

267g of AlCl3 reacted to produce 342g of Al2(SO4)3.

Therefore, 25g of AlCl3 will react to produce = (25 x 342) /267 = 32.02g of Al2(SO4)3.

Therefore, the theoretical yield of Al2(SO4)3 is 32.02g

Step 5:

Determination of the percentage yield of Al2(SO4)3.

This can be obtained as follow:

Actual yield of Al2(SO4)3 = 28.46g

Theoretical yield of Al2(SO4)3 = 32.02g

Percentage yield of Al2(SO4)3 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 28.46/32.02 x 100

Percentage yield = 88.9%

Therefore, the percentage yield of Al2(SO4)3 is 88.9%

3 0
4 years ago
How does an electron emit a photon?
Rudiy27

Answer:

The electron stays in an excited state for a short time. When the electron transits from an excited state to its lower energy state, it will gice off the same amound of energy needed to raise to that level. This emitted energy is a photon.

4 0
3 years ago
Read 2 more answers
You use an analytical balance to measure the mass of a weigh boat and record a mass of 1.5624 g. You then add some NaHCO3 to the
never [62]

Answer:

The minimum number of grams of acetic acid that we would need to fully react with your NaHCO3, is 0.263 grams.

Explanation:

Step 1: Data given

Mass weigh boat = 1.5624 grams

Mass of boat + NaHCO3 = 1.92...

<u>Step 2:</u> Calculate ranges

The minimum possible mass of boat + NaHCO3 = 1.9200g  

Minimum mass of NaHCO3 = 1.9200g - 1.5624g = 0.3576g  

The maximum possible mass of boat + NaHCO3 = 1.9299g  

Maximum mass of NaHCO3 = 1.9299g - 1.5624g = 0.3675g  

What is the minimum mass of acetic acid that must be used to  react with all the NaHCO3 possible in the boat:  

This means what mass of CH3COOH reacts with 0.3675g NaHCO3  

NaHCO3 + CH3COOH → CH3COONa + CO2 + H2O

1 mole of NaHCO3 consumed, needs 1 mole CH3COOH to produce 1 mole of CH3COONa, 1 mole CO2 and 1 mole H2O

<u>Step 3: </u>Calculate moles NaHCO3

Moles NaHCO3 = mass NaHCO3 / Molar mass NaHCO3

moles NaHCO3 = 0.3675 grams / 84g/mol

moles NaHCO3 = 0.004375 moles

Since 1 mole of NaHCO3 consumed, needs 1 mole CH3COOH

For 0.004375 moles NaHCO3, we need 0.004375 moles CH3COOH

<u>Step 4:</u> Calculate mass of CH3COOH

mass of CH3COOH = moles CH3COOH * Molar mass CH3COOH

mass of CH3COOH = 0.004375 moles * 60.05 g/mol

mass CH3COOH = 0.2627 grams ≈ 0.263 grams

The minimum number of grams of acetic acid that we would need to fully react with your NaHCO3, is 0.263 grams.

6 0
3 years ago
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