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lina2011 [118]
3 years ago
8

Naturally occuring boron is 80.20% boron-11 (atomic mass 11.01 amu) and 19.80% of some other isotope. What must the atomic mass

of this second isotope be in order to account for the 10.81 amu average atomic mass of boron?.
Chemistry
1 answer:
Yanka [14]3 years ago
7 0
You will have to do some math. 

<span>80.2 × 11.01 + 19.8 × x = 100 * 10.81. </span>

<span>x is the mass in amu of the second isotope. </span>

<span>Solve for x. </span>

<span>19.8 × x = 100 * 10.81 - 80.2 × 11.01 </span>
<span>19.8 × x = 1081 - 883.00 = 198.00 </span>
<span>x = 1964.00 / 19.8 = 10.00 </span>

<span>The mass of the other isotope is 10.00 amu.</span>
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One mole of ice at 0°C is added to two moles of water at 50°C under a constant external pressure of 1 atm. This process is carri
kotykmax [81]

Answer:

T2 = 29.79°C

Explanation:

Equliibrium signifies that heat loss = heat gained

Heat gained by Ice;

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l = 6.01 k J m o l = 334 J/g

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H =  18(334)

H = 6012

Heat lost by water

H = MCΔT

H = 18 * 4.186 * (50 - T2)

H = 3767.4 - 75.348T2

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6012 = 3767.4 - 75.348T2

- 75.348T2 = 3767 - 6012

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Who preformed the oil drop experiment
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Read 2 more answers
Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . Suppose
Slav-nsk [51]

Answer:

The minimum mass of octane that could be left over is 43.0 grams

Explanation:

Step 1: Data given

Mass of octane = 73.0 grams

Mass of oxygen = 105.0 grams

Molar mass octane = 114.23 g/mol

Molar mass oxygen = 32.0 g/mol

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate the number of moles

Moles = mass / molar mass

Moles octane = 73.0 grams / 114.23 g/mol

Moles octane = 0.639 moles

Moles O2 = 105.0 grams / 32.0 g/mol

Moles O2 = 3.28 moles

Step 4: Calculate the limiting reactant

For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O

O2 is the limiting reactant. It will completely be consumed. (3.28 moles). There will react 3.28 / 12.5 = 0.2624 moles. There will remain 0.639 - 0.2624  = 0.3766 moles octane

Step 5: Calculate mass octane remaining

Mass octane = moles * molar mass

Mass octane = 0.3766 moles * 114.23 g/mol

Mass octane = 43.0 grams

The minimum mass of octane that could be left over is 43.0 grams

3 0
3 years ago
Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa
notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

2NaOH+H_2SO_4-->Na_2SO_4+2H_2O

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4

Finally, the percent yield turns out into:

Y=\frac{1.92g}{7.1g} *100

Y=27.0%

Best regards.

6 0
3 years ago
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