Answer:
<h3>5.) B. 40g</h3><h3>6.) B. About 20g of the KC1O3 Is dissolved</h3>
Explanation:
i hope it helps
<span>a. What is the empirical formula of the gas? In this case, the empirical formula is the same as the molecular formula
</span>
An empirical formula is a formula that gives the proportions of the elements present in a certain compound however it does not give the actual numbers or the arrangement of the atoms. To determine this, we do as follows:
H = (1.69)(.05) = 0.08 g ( 1 mol / 1.01 g ) = 0.0792 / 0.0792 = 1
F = (1.69)(.95) = 1.61 g ( 1 mol / 19 g ) = 0.0847 / 0.0792 = 1
The empirical formula would be HF.
<span>b. What is the molecular formula of the solid?
I bet there is a lacking information for the solid. We cannot determine the molecular formula with only the values given above. The closest would be the compound UF2O2.
</span><span>c. Write a balanced equation for the reaction between UF6 and H2O.
</span>2H2O + UF6 -> UF2O2 +4HF
Answer:
yes
Explanation:
because it is used in electrical conduct.
m not sure
Answer:
79.9 amu
Explanation:
Given data:
Atomic mass of bromine = ?
Percent abundance of 1st isotope = 50.7%
Atomic mass of 1st isotope = 78.92 amu
Percent abundance of 2nd isotope = 49.3%
Atomic mass of 2nd isotope =80.92 amu
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (50.7×78.92)+(49.3×80.92) /100
Average atomic mass = 4001.24 + 3989.36 / 100
Average atomic mass = 7990.6 / 100
Average atomic mass = 79.9 amu.