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trapecia [35]
2 years ago
12

Which of the following is an example of a situation where work occurs?

Chemistry
1 answer:
SIZIF [17.4K]2 years ago
4 0

Answer:

D is the answer may be. I'm not sure.

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What type of bonding is shown in the diagram
Nutka1998 [239]

Answer:

Covalent Bond

Explanation:

In the diagram Carbon and each of the 4 Hydrogens are sharing electrons. They are also both non metals. Both of these are characteristics of Covalent Bonds.

4 0
3 years ago
What category is most of the elements part of
rjkz [21]

Answer:

transition metals

Explanation:

they're the elements in yellow in the picture

6 0
2 years ago
With circulation, the heart provides your body with:
Ivanshal [37]
<h3>Answer:  D) all of the above</h3>

Explanation:

The lungs pump oxygen in and carbon dioxide out, which goes through the blood stream to help with the cell's energy needs.

Nutrients pass through the blood stream as well. The nutrients start with the digestive system (mouth, esophagus, stomach, small intestine) before going into the blood stream. The nutrients are building blocks to help make new cells when old ones die off.

When those cells die off, the body sheds them like dead skin, but internal dead cells are passed off as waste. This waste and other byproducts the body doesn't need passes through the blood stream as well.

In short, the blood stream is basically the highway to help get desired materials (eg: oxygen and nutrients) and get rid of waste (eg: carbon dioxide and other unwanted byproducts or dead cell material)

So that's why the answer includes A, B and C.

8 0
3 years ago
Acetic acid has a Kb of 2.93 °C/m and a normal boiling point of 118.1 °C. What would be the boiling point of a solution made by
alexdok [17]

Boiling point elevation is given as:

ΔTb=iKbm

Where,

ΔTb=elevation in the boiling point

that is given by expression:

ΔTb=Tb (solution) - Tb (pure solvent)

Here Tb (pure solvent)=118.1 °C

i for CaCO3= 2

Kb=2.93 °C/m

m=Molality of CaCO₃:

Molality of CaCO₃=Number of moles of CaCO₃/ Mass of solvent (Kg)

=(Given Mass of CaCO3/Molar mass of CaCO₃)/ Mass of solvent (Kg)

=(100.0÷100 g/mol)/0.4

= 2.5 m

So now putting value of m, i and Kb in the boiling point elevation equation we get:

ΔTb=iKbm

=2×2.93×2.5

=14.65 °C

boiling point of a solution can be calculated:

ΔTb=Tb (solution) - Tb (pure solvent)

14.65=Tb (solution)-118.1

Tb (solution)=118.1+14.65

=132.75

3 0
3 years ago
Help pls , question is in picture
Nata [24]

Explanation:

IM PRETTY SURE IT IS D !! IF ITS WRONG IM SORRY THAT WHAT

I GOT

8 0
3 years ago
Read 2 more answers
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