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2 years ago

Would the amount of solid magnesium increase decrease or stay the same if you added a strong acid to it?

1 answer:

8 0

Solid Magnesium is considered as active metal so it reacts with strong acids like HCl and H₂SO₄ liberating Hydrogen gas according to the following equations:

Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g)
Mg(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂(g)

so the amount of solid magnesium decrease by addition of strong acid to it.

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How much heat energy must be absorbed to completely melt 35.0 grams of H2O(s) at 0°C?

Cerrena [4.2K]

(35/18.02)(6.02kJ)=11.6926 kiloJoules

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3 years ago

a lab is performed to do the study of pH and heat on the function of the catalase enzyme found in carrot plants. what would be c

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It would be the ph and heat

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2 years ago

A solution made by mixing 20.0 g of a non-volatile compound with 125 mL of water at 25°C has a vapor pressure of 22.67 torr. Wha

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We have that the molecular weight (3sf) of the compound (g/mol)

$m=44.15g/mol$

From the question we are told

A solution made by mixing 20.0 g of a non-volatile compound with 125 mL of water at 25°C has a vapor pressure of 22.67 torr. What is the molecular weight (3sf) of the compound (g/mol).

Generally the equation for the Rouault's law is mathematically given as

P=P_0 N

$22.67=23.8*\frac{\frac{12.5}{18}}{\frac{125}{18}+\frac{15}{m}}\\\\\6.95+\frac{15}{m}=7.29\\\\\frac{15}{m}=7.29-6.95\\\\m=\frac{15}{0.34}\\\\m=44.11g/mol$

Therefore

The molecular weight (3sf) of the compound (g/mol)

$m=44.15g/mol$

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2 years ago

A vial containing radioactive selenium-75 has an activity of 3.0 mCi/mL. If 2.6 mCi are required for a leukemia test, how many m

oksian1 [2.3K]

Answer : The $866.66\mu L$ must be administered.

Solution :

As we are given that a vial containing radioactive selenium-75 has an activity of $3.0mCi/mL$ .

As, 3.0 mCi radioactive selenium-75 present in 1 ml

So, 2.6 mCi radioactive selenium-75 present in $\frac{2.6mCi}{3.0mCi}\times 1ml=0.86666ml\times 1000=866.66\mu L$

Conversion :

$(1ml=1000\mu L)$

Therefore, the $866.66\mu L$ must be administered.

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2 years ago

Given the balanced equation, 2 H2 + O2 -> 2 H2O, answer the following

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Answer: $O_2$ is the limiting reagent

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} H_2=\frac{10.0g}{2g/mol}=5.00moles$

$\text{Moles of} O_2=\frac{20.0g}{32g/mol}=0.625moles$

The balanced chemical reaction is :

$2H_2+O_2\rightarrow 2H_2O$

According to stoichiometry :

1 moles of $O_2$ require = 2 moles of $H_2$

Thus 0.625 moles of $O_2$ will require= $\frac{2}{1}\times 0.625=1.25moles$ of $H_2$

Thus $O_2$ is the limiting reagent as it limits the formation of product and $H_2$ is the excess reagent as it is present more than the required amount.

8 0

2 years ago