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il63 [147K]
3 years ago
7

To balance the reaction what coefficients (numbers) are needed: HBr +KOH ---> KBr + H2O

Chemistry
1 answer:
Tom [10]3 years ago
5 0

Answer:

H2Br + 2KOH ----- K2Br + 2H2O

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An element has an average atomic mass of 1.008 amu. It consists of two isotopes , one having a mass of 1.007 amu, and one having
dimulka [17.4K]

Answer:

The most abundant isotope is 1.007 amu.

Explanation:

Given data:

Average atomic mass = 1.008 amu

Mass of first isotope = 1.007 amu

Mass of 2nd isotope = 2.014 amu

Most abundant isotope = ?

Solution:

First of all we will set the fraction for both isotopes

X for the isotopes having mass  2.014 amu

1-x for isotopes having mass 1.007 amu

The average atomic mass is 1.008 amu

we will use the following equation,

2.014x + 1.007  (1-x) = 1.008  

2.014x + 1.007  - 1.007 x = 1.008  

2.014x - 1.007x  =  1.008  -  1.007

1.007 x = 0.001

x= 0.001/ 1.007

x= 0.0009

0.0009 × 100 = 0.09 %

0.09 % is abundance of isotope having mass  2.014 amu because we solve the fraction x.

now we will calculate the abundance of second isotope.

(1-x)

1-0.0009 = 0.9991

0.9991 × 100= 99.91%

6 0
3 years ago
Fe2O3 + 2Al -> Al2O3 + 2Fe
pashok25 [27]
Fe2O3 + 2Al ---> Al2O3 + 2Fe 
Mole ratio Fe2O3 : Al = 1:2 
No. of moles of Fe2O3 = Mass/RMM = 250 / (55.8 * 2 + 16 * 3) = 1.56641604 moles 
No. of moles of Al = 150/27 = 5.555555555 moles. 
Mole ratio 1 : 2. 1.56641604 * 2 = 3.13283208 moles of Al, but you have 5.555555555 moles of Al. So Al is in excess. All of it won't react. 

So take the Fe2O3 and Fe ratio to calculate the mass of iron metal that can be prepared. 
RMM of Fe2O3 / Mass of Fe2O3 = RMM of 2Fe / Mass of Fe 159.6 / 250 = 111.6 / x x = 174.8 g of Fe 
7 0
3 years ago
Read 2 more answers
How many protons and electrons are there in a neutral atom of lithium<br> Np=Ne
pantera1 [17]

Answer:

3 protons and also 3 electrons

Explanation:

z=p=e

3 0
3 years ago
What is the final volume of the solution when 9.80 mL of a 1.56 m solution to need to be diluted to 1.27 m?
musickatia [10]

hAnswer:

Explanation:

8 0
3 years ago
If a 45.6 g piece of aluminum has a density of 2.1 g/mL, what volume would it have?
Mashcka [7]

Answer:

\boxed {\tt 21.7142857 \ mL}

Explanation:

The density formula is:

d=\frac{m}{v}

Rearrange this formula for v, volume. Multiply by v, then divide by d.

d*v=\frac{m}{v}

d*v=m

\frac{d*v}{d} =\frac{m}{d}

v=\frac{m}{d}

Volume can be found by dividing the mass by the density. The mass of the aluminum is 45.6 grams and the density is 2.1 grams per milliliter.

v=\frac{45.6 \ g }{ 2.1 \ g/mL}

Divide. Note the grams, or g, will cancel out.

v=\frac{45.6 }{2.1 \ g}

v= 21.7142857 \ mL

The volume of the aluminum is 21.7142857 milliliters.

3 0
3 years ago
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