Please help with this question and explain too

The mass of proton 1) 1072* 10-28

jasenka [17]

Answer:

Mass of proton is

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5 0

3 years ago

Aspirin (C9H8O4) is synthesized by reacting salicylic acid (C7H6O3) with acetic anhydride (C4H6O3). The balanced equation isC7H6

prisoha [69]

Answer:

For a: The mass of acetic anhydride needed is 73.91 grams.

For b: The theoretical yield of aspirin is 130.43 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 a:

Given mass of salicylic acid = $1.00\times 10^2g=100g$

Molar mass of salicylic acid = 138.121 g/mol

Putting values in equation 1, we get:

$\text{Moles of salicylic acid}=\frac{100g}{138.121g/mol}=0.724mol$

For the given chemical reaction:

$C_7H_6O_3+C_4H_6O_3\rightarrow C_9H_8O_4+CH_3COOH$

By stoichiometry of the reaction:

1 mole of salicylic acid reacts with 1 mole of acetic anhydride.

So, 0.724 moles of salicylic acid will react with = $\frac{1}{1}\times 0.724=0.724mol$ of acetic anhydride.

Now, to calculate the mass of acetic anhydride, we use equation 1:

Moles of acetic anhydride = 0.724 moles

Molar mass of acetic anhydride = 102.09 g/mol

Putting values in equation 1, we get:

$0.724mol=\frac{\text{Mass of acetic anhydride}}{102.09g/mol}\\\\\text{Mass of acetic anhydride}=73.91g$

Hence, the mass of acetic anhydride needed is 73.91 grams.

For b:

By stoichiometry of the reaction:

1 mole of salicylic acid is producing 1 mole of aspirin.

So, 0.724 moles of salicylic acid will produce = $\frac{1}{1}\times 0.724=0.724mol$ of aspirin.

Now, to calculate the mass of aspirin, we use equation 1:

Moles of aspirin = 0.724 moles

Molar mass of aspirin = 180.158 g/mol

Putting values in equation 1, we get:

$0.724mol=\frac{\text{Mass of aspirin}}{180.158g/mol}\\\\\text{Mass of aspirin}=130.43g$

Hence, the theoretical yield of aspirin is 130.43 grams.

4 0

3 years ago

How many grams of sodium sulfide can be produced when 45.3 g Na react with 105 g S?

erastova [34]

Balance the equation: 2Na + S --> Na2S
Using the given amount of the reactants in the reaction, calculate the amount of the product:
45.3g Na x (1 mol/22.99 g)= 1.97 mol of Na
105f S x (1 mol/ 32.06g) = 3.28 mol of S
The limiting reactant would be Na:
1.97 mol Na x (1 mol Na2S/ 2 mol Na) x (78.04g/mol) = 76.87g of Na2S produced

8 0

3 years ago

A rigid tank that contains 2 kg of N2 at 25°C and 550 kPa is connected to another rigid tank that contains 4 kg of O2 at 25°C an

koban [17]

Answer:

The volume in the first tank = 0.32 $m^{3}$

The volume in the second tank = 2.066 $m^{3}$

The final pressure of the mixture = 203.64 K pa

Explanation:

First Tank

Mass = 2 kg

Pressure = 550 k pa

Temperature = 25 °c = 298 K

Gas constant for nitrogen = 0.297 $\frac{KJ}{Kg K}$

From the ideal gas equation

P V = m R T

550 × V = 2 × 0.297 × 298

V = 0.32 $m^{3}$

This is the volume in the first tank.

Second tank

Mass = 4 kg

Pressure = 150 K pa

Temperature = 25 °c = 298 K

Gas constant for oxygen = 0.26 $\frac{KJ}{Kg K}$

From the ideal gas equation

P V = m R T

150 × V = 4 × 0.26 × 298

V = 2.066 $m^{3}$

This is the volume in the second tank.

This is the iso thermal mixing. i.e.

$P_{3} V_{3} = P_{1} V_{1} + P_{2} V_{2}$ ----- (1)

$V_{3} = V_{1} + V_{2}$

$V_{3} = 0.32 + 2.066$

$V_{3} = 2.386 \ m^{3}$

Put this value in equation (1)

$P_{3}$ × 2.386 = 550 × 0.32 + 150 × 2.066

$P_{3}$ = 203.64 K pa

Therefore the final pressure of the mixture = 203.64 K pa

3 0

3 years ago

Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 6.9 g of hexane is mi

MrRa [10]

Answer:

There, there are no leftover for C6H14 instead, an additional mass of 4g of C6H14 is needed to completely react with 38.4g of O2.

Explanation:

Step 1:

We'll begin by writing the balanced equation for the reaction. This is shown below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Step 2:

Let us calculate the masses of C6H14 and O2 that reacted from the balanced equation. This is illustrated below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Molar Mass of C6H14 = (12x6) + (14x1) = 72 + 14 = 86g/mol

Mass of C6H14 from the balanced equation = 2 x 86 = 172g

Molar Mass of O2 = 16x2 =32g/mol

Mass of O2 from the balanced equation = 19 x 32 = 608g

From the balanced equation above, 172g of C6H14 reacted with 608g of O2.

Step 3.

Now, let us determine the mass of C6H14 that will react with 38.4 g of oxygen. This is illustrated below:

From the balanced equation above, 172g of C6H14 reacted with 608g of O2.

Therefore, Xg of C6H14 will react with 38.4g of O2 i.e

Xg of C6H14 = (172 x 38.4) /608

Xg of C6H14 = 10.9g

From the calculations made above, we can see clearly that the mass of C6H14 is limited as the reaction requires 10.9g of C6H14 and only 6.9g was given. There, there are no leftover for C6H14 instead, an additional mass ( 10.9 - 6.9 = 4g) of 4g of C6H14 is needed to completely react with 38.4g of O2.

5 0

2 years ago