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topjm [15]
3 years ago
13

The fact that HBO2, a reactive compound, was produced rather than the relatively inert B2O3 was a factor in the discontinuation

of the investigation of the diborane as a fuel. What mass of liquid oxygen (LOX) would be needed to burn 296.1 g of B2H6?
Chemistry
1 answer:
Reil [10]3 years ago
8 0

Answer:

1027.62 g

Explanation:

For B_2H_6  :-

Mass of B_2H_6  = 296.1 g

Molar mass of B_2H_6  = 27.66 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{296.1\ g}{27.66\ g/mol}

Moles\ of\ B_2H_6= 10.705\ mol

From the balanced reaction:-

B_2H_6(g) + 3 O2_{(l)}\rightarrow 2 HBO_2_{(g)}+ 2 H_2O_{(l)}

1 mole of B_2H_6 react with 3 moles of oxygen

Thus,

10.705 mole of B_2H_6 react with 3*10.705 moles of oxygen

Moles of oxygen = 32.115 moles

Molar mass of oxygen gas = 31.998 g/mol

<u>Mass = Moles * Molar mass = 32.115 * 31.998 g = 1027.62 g</u>

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Eduardwww [97]
The most abundant of all of the isotopes of an element will be the one who's mass the mass of element is closest to. In this case, the mass of atomic carbon is closest to the mass of carbon-12.
Thus, Carbon-12 is the most abundant isotope.
3 0
3 years ago
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A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out
scZoUnD [109]

<u>Answer:</u> The mass percent of potassium bromide in the mixture is 9.996%

<u>Explanation:</u>

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

<u>For lead (II) bromide:</u>

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol

  • The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = \frac{2}{1}\times 0.0021=0.0042mol of potassium bromide

  • Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g

  • To calculate the percentage composition of KBr in the mixture, we use the equation:

\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%

Hence, the percent by mass of KBr in the mixture is 9.996 %

5 0
3 years ago
Give an example of a balanced force.
lutik1710 [3]

Answer:

When two forces are the same strength but act in opposite direction, they are called balanced forces. Again, tug-of-war is a perfect example. If the people on each side of the rope are pulling with the same strength, but in the opposite direction, the forces are balanced. The result is no motion.

Explanation:

Hope this helps!

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I WILL MARK BRAINLIEST PLZ HELP ASAP, thanks :)
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Answer:

Inner

Explanation:

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3 years ago
Individual particles move with the wave.
MArishka [77]

Answer:

No.

Explanation:

No, individual particles do not move with the wave, it only oscillates back and forth its mean position. The particles in the medium transfer its energy to their neighboring particles and in that way the energy moves in the form of wave. The particles only vibrates on its means position instead of moving from one place to another. So we can conclude that Individual particles do not move with the wave.

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