The answer is 267.93 g
Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol
The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%
Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g
Answer:
- Take 3.3 mL of 3.0-M hydrochloric acid and subsequently add 76.7 mL of water to complete the 100.00 mL.
- Take 11.7mL of 6.0-M hydrochloric acid and subsequently add 88.3 mL of water to complete the 100.00 mL
Explanation:
Hello,
In this case, given that the dilutions are preparedfrom 3.0-M and 6.0-M hydrochloric acid, we must proceed as follows:
- 3.0-M stock: when using this stock, the aliquot you must take is computed as shown below:

It means that you must take 23.3 mL of 3.0-M hydrochloric acid and subsequently add 76.7 mL of water to complete the 100.00 mL.
- 6.0-M stock: when using this stock, the aliquot you must take is computed as shown below:

It means that you must take 11.7mL of 6.0-M hydrochloric acid and subsequently add 88.3 mL of water to complete the 100.00 mL.
Regards.
3x10^8 / (670.8 * 10^-9) =4.47x10^14 Hz 4.47x10^14 Hz multiplied by plank's constant = 2.9634x10^-19
2.96 x10-19 J
Question 1 im pretty sure it is <span>a. oxygen is in the formula.
</span>question 2 is b. dinitrogen pentoxide
1st you pour the mixture<span> through a filter to remove the larger </span>pebbles<span>. Next, add water to dissolve the salt and then filter out the </span>pepper the last thing you do <span>evaporate the water to leave the salt behind.</span>