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Serga [27]
3 years ago
9

How many grams are in 1.6 moles of potassium bromide?​

Chemistry
1 answer:
slava [35]3 years ago
6 0

Answer:

190.4g

Explanation:

1.6mol of KBr (119.002g KBr/1 mol) = 190.4g

since you want to find grams, take the molar mass of KBr (119.002) per 1 mol and use it as your conversion factor (119.002g KBr/1 mol) which will then cancel out mols and leave you with grams.

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Why is the solar system often referred to as the sun’s family ?
snow_tiger [21]

Answer:the suns family is stars think about how many stars are in the sky there are billons and more billions

Explanation:

7 0
3 years ago
Read 2 more answers
What is a control group used for?
jeka94
  • The control group defined as  a group that does not include any change to the inconstant being tested. Why is a control crucial in an experiment? The control group is crucial because it acts as a benchmark to variatethe results of the experiment to. The experimental group is the group that the scientist is examing . The experimental group receives a change to a variable, or the conditions allowed to change in the experiment.
  • There are two types of control group :   Positive control groups and Negative control groups
  • The positive control group is a group that is contrive to produce the effect you are looking for in the experimental group. The positive control group shows the scientists that the craved results are achievable . This helps stop false negative outcome in the experimental group, where a negative result is receive but is due to a failure in the experiment instead than a truly negative result based on the experimental conditions.
  • A negative control group is a group that is not exposed to the different  scientist is testing, called the independent variable. A negative control group function as a benchmark to secure that the results that are got are actually due to the independent variable and not anything else

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3 0
1 year ago
If the half-life of a radioactive substance is 2 seconds, how much of a 100-gram sample would be left after 8 seconds?
Elan Coil [88]
M₀=100 g
T=2 s
t=8s

m=m₀/{2^(t/T)}

m=100/{2^(8/2)}=100/2⁴=6.25 g
4 0
3 years ago
Solve the equation:<br> p - 15 = 73<br> p = [?]
GenaCL600 [577]

Answer:

88

Explanation:

p-15=73

p=73+15

p=88

5 0
3 years ago
A 8.249 gram sample of copper is heated in the presence of excess fluorine. A metal fluoride is formed with a mass of 13.18 g. D
levacccp [35]

Answer:

CuF_2 the empirical formula of the metal fluoride.

Explanation:

Mass of copper heated = 8.249 g

Mass of copper fluoride formed = 13.18 g

Mass of fluorine gas in copper fluoride = x

13.18 g = 8.249 g + x\\x= 13.18 - 8.249 g = 4.931 g

Moles of copper :

= \frac{8.249 g}{63.546 g/mol}=0.1298 mol

Moles of fluorine:

= \frac{4.931 g}{18.998 g/mol}=0.2596 mol

For the empirical formula divide the smallest mole of an element with all the moles of elements present in the compound.

Copper= \frac{0.1298 mol}{0.1298 mol}=1\\Fluorine = \frac{0.2596 mol}{0.1298 mol}=2

The empirical formula of the copper fluoride = CuF_2

CuF_2 the empirical formula of the metal fluoride.

6 0
3 years ago
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