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saveliy_v [14]
3 years ago
5

Frayer model for matter

Chemistry
1 answer:
givi [52]3 years ago
6 0
A model showing that gases are made from the matter of particles that are too small to see and are moving freely around in space can explain many observations.
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Billy the friendly Robot uses 50 N of force to lift a box 2 meters in the air. How much work did Billy perform?
nexus9112 [7]

Answer:

Work done, W = 100 J

Explanation:

We have, Billy the friendly Robot uses 50 N of force to lift a box 2 meters in the air.

It is required to find the work done by Billy.

Work done by an object is given in terms of force and displacement. The formula used to find the work done is given by :

W=Fd\\\\W=50\ N\times 2\ m\\\\W=100\ J

So, the work performed by Billy is 100 J.

4 0
3 years ago
Help I need this lol!!!!!
brilliants [131]

Answer:

green

Explanation:

4 0
3 years ago
Not all of the water falls as rain, snow,or sleet evaporates. What happens to the rest of the water?
Bumek [7]
Hi there! All of the water returns to the atmosphere, it's called evapotranspiration and condenses. And then the water falls back to the Earth's surface and starts the cycle all over again. If you don't know what I mean, it's the Water Cycle.
4 0
3 years ago
Which atomic property is different in each isotope of an element?
Valentin [98]
Every isotope of an element has a different number of neutrons, which means that the atomic property which is different in each isotope of an element is mass number.
Mass number depends on the number of neutrons in an element.
6 0
3 years ago
Based on the equation, how many grams of Br2 are required to react completely with 72.4 grams of AlCl3
svlad2 [7]

Answer:  131 g of bromine is required.

Explanation:

The balanced equation will be :

2AlCl_3+3Br_2\rightarrow 2AlBr_3+3Cl_2

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

moles of AlCl_3

\text{Number of moles}=\frac{72.4g}{133g/mol}=0.544moles

According to stoichiometry :

2 moles of AlCl_3 require  = 3 moles of Br_2

Thus 0.544 moles of AlCl_3 require=\frac{3}{2}\times 0.544=0.816moles  of Br_2

Mass of Br_2=moles\times {\text {Molar mass}}=0.816moles\times 160g/mol=131g

Thus 131 g of bromine is required.

5 0
3 years ago
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