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3 years ago

HELP PLEASE!!!!! show work

Chemistry

1 answer:

muminat3 years ago

5 0

I’m not sure how to show the work
You should download the app chem balance
It’s great for this kind of homework

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A potassium bromide solution is 8.30% potassium bromide by mass and its density is 1.03 g/ml. what mass of potassium bromide is

pshichka [43]

<u>Answer:</u> The mass of potassium bromide present in 41.2 mL of solution will be 3.522 grams.

<u>Explanation:</u>

We are given that KBr is present in 8.3% KBr solution, which means that 8.3 grams of potassium bromide is present in 100 gram of the solution.

To calculate the volume of KBr, we use the formula:

$Density=\frac{Mass}{Volume}$

Mass of the solution = 100 grams

Density of KBr solution = 1.03g/mL

Volume of the solution = ? mL

Putting values in above equation, we get:

$1.03g/mL=\frac{100g}{Volume}\\\\Volume=97.08mL$

Now, to calculate the mass of KBr in 41.2mL of the solution, we use unitary method.

In 97.08 mL of solution, mass of KBr present is 8.3 grams.

So, 41.2 mL of solution will contain = $\frac{8.3g}{97.08mL}\times 41.2mL=3.522g$ of KBr.

Hence, the mass of potassium bromide present in 41.2 mL of solution will be 3.522 grams.

7 0

3 years ago

What element is represented by [Rn]7s1?

lara [203]

Answer:

Radon

Explanation:

4 0

3 years ago

What coefficient must be added to the carbon in the reactants to balance the following equation?

Sophie [7]

Answer:

Explanation:

There are 3 carbons on the right side

there are only two on the left side....you need one more ...so add one more... change the '1' coefficient in front of C to '2'

7 0

2 years ago

PLEASE HELP ASAP!!!!!!!!

MaRussiya [10]

Answer:

You need to count all the atoms on each side of the chemical equation. once you know how many of each type of atom you have,you can only change the coefficient. (the numbers in front of the atoms or compounds

7 0

2 years ago

On a mission to a newly discovered planet, an astronaut finds chlorine abundances of 13.85 % for 35Cl and 86.15 % for 37Cl. What

diamong [38]

Answer:

36.693

Explanation:

From the question given, the following were obtained:

Let 35Cl be isotope A and 37Cl be isotope

For isotope A (35Cl),

Mass number = 34.9700

Abundance = 13.85 %

For isotope B (37Cl),

Mass number = 36.9700

Abundance = 86.15 %

Atomic mass =

[( Mass of AxAbundance A)/100] + [(Mass of BxAbundance B)/100]

[(34.97x13.85)/100] + [(36.97x86.15)/100]

= 36.693

6 0

2 years ago