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3 years ago

Please Help

1 answer:

5 0

I believe it is A because a physical change is something that can be changed back to where a chemical change can not be such as burning something.

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If you burn 46.0 g of hydrogen and produce 411 g of water, how much oxygen reacted?

g100num [7]

The answer is 365 grams because you need to subtract 411 by 46.

3 0

3 years ago

12.A piece of magnesium is in the shape of a cylinder with a height of 5.62 cm

yaroslaw [1]

Answer:

Density, d = 1.779 g/cm³

Explanation:

The density of a material is given by its mass per unit volume.

Here, height of a piece of magnesium cylinder, h = 5.62 cm

Its diameter, d = 1.34 cm

Radius = 0.67 cm

Volume of he cylinder,

$V=\pi r^2 h\\\\\text{Putting the value of r and h, we get :}\\\\V=(\pi \times (0.67)^2\times 5.62)\ cm^3$

$d=\dfrac{m}{V}\\\\d=\dfrac{14.1\ g}{(\pi \times (0.67)^2\times 5.62)\ cm^3}\\\\d=1.779\ g/cm^3$

So, the density of the sample is 1.779 g/cm³.

4 0

3 years ago

Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K

Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

$MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-$

And the undergoing chemical reaction:

$MgCl_2+2NaF\rightarrow MgF_2+2NaCl$

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

$n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2$

Next, the moles of magnesium chloride consumed by the sodium fluoride:

$n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2$

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

$n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2$

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

$[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M$

$[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M$

Thereby, the reaction quotient is:

$Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}$

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0

3 years ago

Which of these is the final result of secondadry succession?

Grace [21]

Answer:

Explanation:

ADAPTATIOnn but if thhere would be an option o all the above it would be that

5 0

3 years ago

Read 2 more answers

If you try to balance an equation by changing subscripts you change...

aksik [14]

Answer:

kwkrofofoxosowoqoaododpdprofpcoxozoskawkdjdn

Explanation:

sklwlrlfclxoskkekrdododosoekekrkrododowoekekfkdodkwkeororkdkdkwejrjrkfidiwi3jr

8 0

2 years ago