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lilavasa [31]
3 years ago
8

What kinds of things could a magnet make another object or magnet do without touching it?​

Chemistry
1 answer:
Sidana [21]3 years ago
8 0

Answer:

Magnets can be used to make other magnets and things made of iron move without being touched. Something that has been electrically charged can make other things move without touching them.

Explanation:

(can someone help me with my question?)

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Which of the following is not apart of daltons atomic theory
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3. Atoms change into other atoms in chemical reactions.

The idea that you can transform one element into another was called alchemy, and was proven untrue a long time ago. 
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Two of the simplest compounds containing just carbon and hydrogen are methane and ethane. Methane contains 0.3357 g of hydrogen
Natasha_Volkova [10]
Let the ratio of grams of hydrogen per gram of carbon in methane be M, we know that:
M = 0.3357 g / 1 g

Next, lets represent the grams of hydrogen per gram of carbon in ethane be E. The final piece of information we have is:

M / E = 4/3

If we cross multiply,

3M = 4E

Now, substituting the value of M from earlier and solving for E,

E = (3 * 0.3357) / 4
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There are 0.2518 grams of hydrogen per gram of carbon in ethane.
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What is the difference between applied and pure chemistry?
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3 years ago
Calculate the molar solubility of CaF2 in a 0.25 m solution of NaF(aq).
Kipish [7]

Answer:

6.4 × 10^-10 M

Explanation:

The molar solubility of the ions in a compound can be calculated from the Ksp (solubility constant).

CaF2 will dissociate as follows:

CaF2 ⇌Ca2+ + 2F-

1 mole of Calcium ion (x)

2 moles of fluorine ion (2x)

NaF will also dissociate as follows:

NaF ⇌ Na+ + F-

Where Na+ = 0.25M

F- = 0.25M

The total concentration of fluoride ion in the solution is (2x + 0.25M), however, due to common ion effect i.e. 2x<0.25, 2x can be neglected. This means that concentration of fluoride ion will be 0.25M

Ksp = {Ca2+}{F-}^2

Ksp = {x}{0.25}^2

4.0 × 10^-11 = 0.25^2 × x

4.0 × 10^-11 = 0.0625x

x = 4.0 × 10^-11 ÷ 6.25 × 10^-2

x = 4/6.25 × 10^ (-11+2)

x = 0.64 × 10^-9

x = 6.4 × 10^-10

Therefore, the molar solubility of CaF2 in NaF solution is 6.4 × 10^-10M

8 0
3 years ago
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