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Georgia [21]
3 years ago
8

PLEASE READ MARKING BRAINIEST

Chemistry
1 answer:
balandron [24]3 years ago
7 0

Answer:

MgF2 + Li2CO3 -> MgCO3 + 2LiF

It is important to balance chemical equations because there must be an equal number of atoms on both sides of the equation to follow the Law of the Conservation of Mass

Write down the element and each amount on each side of the equation

Then compare each side and add your coefficients so the amounts of each element are equal

Explanation:

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Sound _____.
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Answer:

Sounds travels in transverse waves requires a medium to travel through

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2 years ago
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Give an example of an element that would be expected to share similar chemical properties with beryllium, calcium, and strontium
Aleks04 [339]
Strontium possibly but I’m not 100% sure
4 0
3 years ago
A student is adding DI water to a volumetric flask to make a 50% solution. Unfortunately, he was not paying attention and filled
dalvyx [7]

Answer:

His results will be skewed because there was more water than stock solution. Which would cause the percentage solution to be less than 50% therefore the density would be less than the actual value.

Explanation:

The solution will have percentage less than that of 50%. Therefore the density would be less than the actual value.

Suppose there should be 50 mL of the solution, and he added 60 mL. So 10 mL of the solution is added more.

Suppose the mass of the solute is m.

Originally, the density is = $\frac{m}{50}$     \left(\frac{\text{mass}}{\text{volume}}\right)

Now after adding extra 10 mL , the density becomes $\frac{m}{60}$.

Therefore, $\frac{m}{50}>\frac{m}{60}$

So the density decreases when we add more solution.

4 0
3 years ago
Chemistry: counting atoms in compounds worksheet #7.0.1<br> Answer please!!!!!
stealth61 [152]
Counting atoms in a compound can be done by taking one element at a time and multiplying the subscript of the element and the number of molecules of the compounds. For example, H2O, there are two atoms of H adn 1 atom of oxygen. 
7 0
3 years ago
g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide sol
Ronch [10]

Answer:

M=0.0637M

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2}  *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}

Finally, the resulting molarity in 30.8 mL (0.0308 L):

M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M

Regards.

3 0
2 years ago
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