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allochka39001 [22]
3 years ago
15

What is the percent composition of Copper in CuBr2?

Chemistry
1 answer:
Luda [366]3 years ago
6 0

The answer is 28% as u find the molar mass of cu and divide it by the total molar mass

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"a basic experiment involves a minimum of ________ participant group(s)."
3241004551 [841]
Answer: TWO.

At least two groups: treatment group and control group.

The treatment group is that is exposed to the different levels of the independent variable ( a medication for example), while the control group is not treated, so the researchers can compare the effect of the medication.
5 0
2 years ago
How many moles of SnF₂ will be produced along with 48 grams of H₂? *
KIM [24]

Answer: 24 moles of SnF_2 are produced.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

{\text{Moles of}  H_2}=\frac{48g}{2g/mol}=24moles

Sn+2HF\rightarrow SnF_2+H_2

According to stoichiometry :

1 mole of H_2 is accompanied with = 1 mole of SnF_2

Thus 24 moles of H_2  is accompanied with =\frac{1}{1}\times 24=24moles  of SnF_2

Thus 24 moles of SnF_2 are produced.

8 0
2 years ago
What is the molar mass for FeBr3
AleksAgata [21]
Mass of Iron (Fe): 55.845Mass of Bromine (Br): 79.904
You need to multiply the mass of Br by 3 because there are 3 Bromine atoms. 
(79.904)(3)+ 55.845= 239.712+55.845                                = 295.557 g/mol

6 0
2 years ago
What would be the mass of 9.03*10^21 molecules of hydrobromic acid
Fynjy0 [20]

Answer:

2.11 g hydrobromic acid (correct to 3SF)

Explanation:

Molecular formula of hydrobromic acid = C2H5BrO2

mass of C2H5BrO2 = 140.96g

Beginning with what we're given, 9.03*10^21 we then make a conversion by using Avegadro's number which is 6.02*10^23 per mole (Oct. 23 at 6:02 am is national mole day :) Then, we need to convert out of moles, 140.96g hydrombromic acid per mole.

It looks like this:

9.03*10^21 molecules • (1 mol C2H5BrO2 / 6.02*10^23 molecules) • (140g C2H5BrO2 / 1 mol) = 2.1144 g C2H5BrO2

3 0
3 years ago
Complete this neutralization equation:<br> H2SO4 + Al(OH)3
elena-s [515]

Answer:

H2SO4 + Al(OH)3 = Al2(SO4)3 + H2O

Explanation:

4 0
3 years ago
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