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3 years ago

10

When magnesium loses it's valence electrons, it has the atomic structure of what element?

1 answer:

Arte-miy333 [17]3 years ago

4 0

As far as I can tell the best answer for this would be (A) Neon. However, I would argue that this is at the very least a misleading question. Atoms are less identified by their electrons than their protons (which is represented always by its atomic number). Although atoms can gain or lose electrons, the protons would never change (and remain the same element). Personally, I would have written the question as, "When Magnesium loses its valence electrons, its new number of electrons would most closely resemble _____"

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a gas measured a volume 100 mL under pressure of 740 mmHg. What would the volume be under a pressure of 780 mmHg with constant t

nikdorinn [45]

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<h2>94.87 mL</h2>

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The new volume can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we're finding the new volume

$V_2 = \frac{P_1V_1}{P_2} \\$

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<h3>94.87 mL</h3>

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2 years ago

A gas mixture contains SO 2 SO2 ( molar mass = 64 g/mol ) (molar mass=64 g/mol) and no other source of sulfur. If the mixture is

Slav-nsk [51]

Answer : The percentage (by mass) of $SO_2$ in the mixture is 20 %

Explanation :

As we are given that, 10 % sulfur by mass that means 10 grams of sulfur present in 100 grams of mixture.

Mass of sulfur = 10 g

Mass of mixture = 100 g

Now we have to calculate the mass of $SO_2$ .

As, 32 grams of sulfur present in 64 grams of $SO_2$

So, 10 grams of sulfur present in $\frac{10}{32}\times 64=20$ grams of $SO_2$

Thus, the mass of $SO_2$ is 20 grams.

Now we have to calculate the percentage (by mass) of $SO_2$ in the mixture.

$\% \text{ of }SO_2=\frac{\text{Mass of }SO_2}{\text{Mass of mixture}}\times 100$

$\% \text{ of }SO_2=\frac{20g}{100g}\times 100=20\%$

Therefore, the percentage (by mass) of $SO_2$ in the mixture is 20 %

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2 years ago