Which statement is true?

How many milliners of hydrogen gas ar produced by the reaction 0.020 moles of magnesium with excess of hydrochloride acid at sto

jeyben [28]

Answer:- 448 mL of hydrogen gas are formed.

Solution:- It asks to calculate the volume of hydrogen gas formed in milliliters at STP when 0.020 moles of magnesium reacts with excess HCl acid. The balanced equation is:

$Mg+2HCl\rightarrow MgCl_2+H_2(g)$

There is 1:1 mol ratio between Mg and hydrogen gas. So, the moles of hydrogen gas is also equals to the moles of Mg reacted.

moles of Hydrogen gas formed = 0.020 mol

At STP, volume of 1 mol of the gas is 22.4 L. We need to calculate the volume of 0.02 moles of hydrogen gas.

$0.02mol(\frac{22.4L}{1mol})$

= 0.448 L

They want answer in mL. So, let's convert L to mL using the conversion formula, 1L = 1000mL

$0.448L(\frac{1000mL}{1L})$

= 448 mL

So, 0.020 moles of magnesium would produce 448 mL of hydrogen gas at STP on reacting with excess of HCl acid.

5 0

3 years ago

Calculate the atomic mass of silver if 13 out of 25 atoms are silver-107 and 12 out of 25 are silver-109

nekit [7.7K]

From,

RAM=element×its relative abudance/total abudance

=((107×13)+(12×109))/25

The answer is=107.96

3 0

3 years ago

A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?

Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

$\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles$

Now we have to calculate the moles of $Br_2$

${\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles$

${\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles$

The balanced chemical reaction is,

$2Na(s)+Br_2(g)\rightarrow 2NaBr$

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium bromide

Now we have to calculate the percent yield of reaction

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%$

Therefore, the percent yield is, 93.4%

3 0

3 years ago

In which mutation does one base take the place of another?

Neporo4naja [7]

Answer: A substitution is a mutation that exchanges one base for another (i.e., a change in a single "chemical letter" such as switching an A to a G). Such a substitution could: change a codon to one that encodes a different amino acid and cause a small change in the protein produced.

Explanation: I know this is correct just trust me and please mark me as brainiest.

4 0

3 years ago

There are two naturally occurring isotopes of indium: indium-113 and indium-115. How many neutrons are in a single atom of indiu

KonstantinChe [14]

There are 66 neutrons in a single atom of indium-115. The atomic number of indium-115 is 49, meaning there are 49 protons. Then the atomic mass is 115, so 115-49 = 66.

5 0

3 years ago