How many grams of N2 are in 44.8L at STP?

If you have 110.0 grams of an unknown compound that contains 12.3 grams of hydrogen, what is the percent by mass of hydrogen in

Vladimir [108]

All you have to do is a simple division "parts over the whole"

12.3 grams H/ 110 grams compound x 100= 11.2%

8 0

3 years ago

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What is the speed of a cheetah that runs 30 miles in 0.5 hours? a

REY [17]

Answer:

60 mph (miles per hour)

Explanation:

0.5 hours is 1/2 of an hour, so to get the number of miles for a whole hour you multiply the miles ran by 2.

30 times 2 is 60.

6 0

2 years ago

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How does carbon dioxide get into the atmosphere?

e-lub [12.9K]

Answer:

By condensation.

Explanation:

Have a nice day dear

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3 years ago

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The volume of a sample gas, initially at 25 C and 158 mL, increased to 450 mL. What is the final temperature of the sample of ga

Rashid [163]

Answer:

Final temperature of the gas is 576 $^{0}\textrm{C}$ .

Explanation:

As the amount of gas and pressure of the gas remains constant therefore in accordance with Charles's law:

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

where $V_{1}$ and $V_{2}$ are volume of gas at $T_{1}$ and $T_{2}$ temperature (in kelvin scale) respectively.

Here $V_{1}=158mL$ , $T_{1}=(273+25)K=298K$ and $V_{2}=450mL$

So $T_{2}=\frac{V_{2}T_{1}}{V_{1}}=\frac{(450mL)\times (298K)}{(158mL)}=849K$

849 K = (849-273) $^{0}\textrm{C}$ = 576 $^{0}\textrm{C}$

So final temperature of the gas is 576 $^{0}\textrm{C}$ .

3 0

3 years ago

The expression of the theoretical yield (TY) in function of limiting reagent (LR) of a reaction is as follows: TY = ideal mole r

spin [16.1K]

Answer: The theoretical yield of acetanilide is 6.5 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For aniline:

Given mass of aniline = $4.50\times 10^0=4.50g$ (We know that: $10^0=1$ )

Molar mass of aniline = 93.13 g/mol

Putting values in equation 1, we get:

$\text{Moles of aniline}=\frac{4.50g}{93.13g/mol}=0.048mol$

For acetic anhydride:

To calculate the mass of acetic anhydride, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Volume of acetic anhydride = $(1.25\times \text{Mass of aniline})=1.25\times 4.50=5.625mL$

Density of acetic anhydride = 1.08 g/mL

Putting values in above equation:

$1.08g/mL=\frac{\text{Mass of acetic anhydride}}{5.625mL}\\\\\text{Mass of acetic anhydride}=(1.08g/mL\times 5.625mL)=6.08g$

Given mass of acetic anhydride = 6.08 g

Molar mass of acetic anhydride = 102.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of acetic anhydride}=\frac{6.08g}{102.1g/mol}=0.06mol$

The chemical equation for the reaction of aniline and acetic anhydride follows:

$C_6H_5NH_2+CH_3COOCOCH_3\rightarrow C_6H_5NHCOCH_3+CH_3COOH$

By Stoichiometry of the reaction:

1 mole of aniline reacts with 1 mole of acetic anhydride

So, 0.048 moles of aniline will react with = $\frac{1}{1}\times 0.048=0.048mol$ of acetic anhydride

As, given amount of acetic anhydride is more than the required amount. So, it is considered as an excess reagent.

Thus, aniline is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of aniline produces 1 mole of acetanilide

So, 0.048 moles of aniline will produce = $\frac{1}{1}\times 0.048=0.048mol$ of acetanilide

Now, calculating the theoretical yield of acetanilide by using equation 1:

Moles of acetanilide = 0.048 moles

Molar mass of acetanilide = 135.17 g/mol

Putting values in equation 1, we get:

$0.048mol=\frac{\text{Mass of acetanilide}}{135.17g/mol}\\\\\text{Mass of acetanilide}=(0.048mol\times 135.17g/mol)=6.5g$

Hence, the theoretical yield of acetanilide is 6.5 grams.

3 0

3 years ago