How many grams of sucrose must be added to 375 mL of watertoprepare a 2.75m/m percent solution of sucrose?

Answers

Answer 1

Answer:

Answer : The mass of sucrose added to 375 mL of water must be, 10.6 grams.

Explanation :

As we are given that 2.75 m/m percent solution of sucrose. That means, 2.75 grams of sucrose present in 100 grams of solution.

Mass of solution = 100 g

Mass of sucrose = 2.75 g

Mass of water = Mass of solution - Mass of sucrose

Mass of water = 100 g - 2.75 g

Mass of water = 97.25 g

First we have to calculate the mass of water.

$\text{Mass of water}=\text{Density of water}* \text{Volume of water}$

Density of water = 1.00 g/mL

Volume of water = 375 mL

$\text{Mass of water}=1.00g/mL* 375mL=375g$

Now we have to calculate the mass of sucrose in 375 g of water.

As, 97.25 grams of water contain 2.75 grams of sucrose

So, 375 grams of water contain $(375)/(97.25)* 2.75=10.6$ grams of sucrose

Therefore, the mass of sucrose added to 375 mL of water must be, 10.6 grams.

Answer 2

Answer:

To make a 2.75% m/m sucrose solution, you need to add approximately 1062 grams of sucrose to 375 mL of water, considering the density of water as 1 g/mL.

To prepare a mass/mass (m/m) percent solution of sucrose, you need to calculate the mass of sucrose (in grams) that needs to be added to 375 mL of water to achieve a 2.75% concentration.

Here's how you can calculate it:

1. Convert the volume of water to grams, considering the density of water:

Density of water ≈ 1 g/mL

Mass of water = Volume of water × Density of water

Mass of water = 375 mL × 1 g/mL = 375 g

2. Determine the desired mass of sucrose as a percentage of the total mass:

Desired m/m percent = 2.75%

3. Calculate the mass of sucrose needed:

Mass of sucrose = (Desired m/m percent / 100) × Total mass

Mass of sucrose = (2.75 / 100) × (375 g + Mass of sucrose)

4. Rearrange the equation to solve for the mass of sucrose:

Mass of sucrose = (2.75 / 100) × (375 g) / (1 - (2.75 / 100))

Now, calculate:

Mass of sucrose = (2.75 / 100) × (375 g) / (1 - 0.0275)

Mass of sucrose ≈ (2.75 / 100) × (375 g) / 0.9725

Mass of sucrose ≈ (2.75 × 375 g) / 100 / 0.9725

Mass of sucrose ≈ (1031.25 g) / 0.9725

Mass of sucrose ≈ 1061.98 g

So, approximately 1062 grams of sucrose must be added to 375 mL of water to prepare a 2.75 m/m percent solution of sucrose.

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The maximum allowable concentrationof lead in drinking water is 9.0 ppb. If 2.0
X 10 grams of lead is present in 250
mL of water, is it safe to drink the water?
Support your answer with mathematical
proof.

Answers

We are given:

Maximum concentration of Lead in water = 9 ppb

Mass of Lead in the given solution = 20 grams

Volume of water in the given solution = 250 mL

What is 1 ppb?

1 ppb, short for parts-per-billion. As from its name itself, ppb is used to find how many molecules of solute are present per 1 Billion molecules of the solvent

you can also use it in grams to get the formula:

1 ppb = 1 gram of solute / 1 Billion grams of Solvent

Finding the Ideal ppb concentration:

We are given that the maximum allowed concentration is 9 ppb

which means that we need 9 grams of the solute per 1 Billion grams of Solvent: 9 grams of Solute / 10⁹ grams of Solvent

ppb Concentration of the given solution:

We have 20 grams of Solute in 250 mL of water

Since the density of water is 1 gram/mL

20 grams of Solute / 250 grams of Solvent

As we can see, this fraction is FAR more large that the maximum ppb concentration

This means that the concentration of Lead in the Given solution is higher than the maximum amount and Hence, is unfit to drink

Indicate the number of protons, neutrons, and electrons in each of the following species: 15N7, 33S16, 63Cu29, 84Sr38, 130Ba56, 186W74, 202Hg80

Answers

The number of protons neutrons, and electrons in each of the following species given are below;

What is the atomic number?

The total number of protons present in an atom is known as the atomic number of that atom. The atomic number has no correlation either with the number of neutrons or the number of electrons present inside an atom.

15N7 ⇒ 7 electrons, 8 neutrons, 7 protons

33S16 ⇒ 16 protons, 16 electrons, 17 neutrons

63Cu29 ⇒ 29 electrons, 34 neutrons,29 protons

84Sr38 ⇒ 38 electrons, 46 neutrons,38 protons

130Ba56 ⇒ 56 electrons, 74 neutrons,56 protons

186W74⇒ 74 electrons, 112 neutrons,74 protons

202Hg80 ⇒ 80 electrons, 122 neutrons ,80 protons

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Answer:

1. 7 protons, 7 electrons, 8 neutrons

2. 16 protons, 16 electrons, 17 neutrons

3. 29 protons, 29 electrons, 34 neutrons

4. 38 protons, 38 electrons, 46 neutrons

5. 56 protons, 56 electrons, 74 neutrons

6. 74 protons, 74 electrons, 112 neutrons

7. 80 protons, 80 electrons, 122 neutrons

State and explain grahms law of diffusion

Answers

Answer:

Graham's law of effusion was formulated by Scottish physical chemist Thomas Graham in 1848. Graham found experimentally that the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles. This formula can be written as: , where: Rate₁ is the rate of effusion for the first gas.

Explanation:

r1 = rate of effusion for gas 1

r2 = rate of effusion for gas 2

M1 = molar mass of gas 1

M2 = molar mass of gas 2

Consider the balanced equation for the following reaction:7O2(g) + 2C2H6(g) → 4CO2(g) + 6H2O(l)
Determine the amount of CO2(g) formed in the reaction if 8.00 grams of O2(g) reacts with an excess of C2H6(g) and the percent yield of CO2(g) is 90.0%.

Answers

Answer: The amount of carbon dioxide formed in the reaction is 5.663 grams

Explanation:

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

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

Given mass of oxygen gas = 8 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=(8g)/(32g/mol)=0.25mol$

For the given chemical equation:

$7O_2(g)+2C_2H_6(g)\rightarrow 4CO_2(g)+6H_2O(l)$

By Stoichiometry of the reaction:

7 moles of oxygen gas produces 4 moles of carbon dioxide

So, 0.25 moles of oxygen gas will produce = $(4)/(7)* 0.25=0.143mol$ of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.143 moles

Putting values in equation 1, we get:

$0.143mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\n\n\text{Mass of carbon dioxide}=(0.143mol* 44g/mol)=6.292g$

To calculate the experimental yield of carbon dioxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100$

Percentage yield of carbon dioxide = 90 %

Theoretical yield of carbon dioxide = 6.292 g

Putting values in above equation, we get:

$90=\frac{\text{Experimental yield of carbon dioxide}}{6.292g}* 100\n\n\text{Experimental yield of carbon dioxide}=(90* 6.292)/(100)=5.663g$

Hence, the amount of carbon dioxide formed in the reaction is 5.663 grams

What is the mass in grams of H₂ that can be formed from 54.6 grams of NH₃ in the following reaction?2 NH₃(g) → 3 H₂(g) + N₂(g)

Answers

9.6 grams of H₂ can be formed from 54.6 grams of NH₃ in the following reaction: 2NH₃(g) → 3H₂(g) + N₂(g).

According to this question, the following balanced equation is given: 2NH₃(g) → 3H₂(g) + N₂(g).

First, we convert the mass of ammonia (NH3) to moles as follows:

moles of NH3 = 54.6g ÷ 17g/mol

moles of NH3 = 3.2mol.

If 2 moles of NH3 produces 3 moles of H2.
3.2 moles of NH3 will produce 4.8 moles of H2.

Next, we convert 4.8moles of H2 to mass as follows:

mass of H2 = 4.8 × 2

mass of H2 = 9.6g of H2.

Therefore, 9.6 grams of H₂ can be formed from 54.6 grams of NH₃ in the following reaction: 2NH₃(g) → 3H₂(g) + N₂(g).

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Answer : The mass of $H_2$ is, 9.64 grams.

Explanation : Given,

Mass of $NH_3$ = 54.6 g

Molar mass of $NH_3$ = 17 g/mol

Molar mass of $H_2$ = 2 g/mol

First we have to calculate the moles of $NH_3$ .

$\text{Moles of }NH_3=\frac{\text{Given mass }NH_3}{\text{Molar mass }NH_3}$

$\text{Moles of }NH_3=(54.6g)/(17g/mol)=3.21mol$

Now we have to calculate the moles of $H_2$

The balanced chemical equation is:

$2NH_3(g)\rightarrow 3H_2(g)+N_2(g)$

From the balanced reaction we conclude that

As, 2 mole of $NH_3$ react to give 3 moles of $H_2$

So, 3.21 mole of $NH_3$ react to give $(3)/(2)* 3.21=4.82$ mole of $H_2$

Now we have to calculate the mass of $H_2$

$\text{ Mass of }H_2=\text{ Moles of }H_2* \text{ Molar mass of }H_2$

$\text{ Mass of }H_2=(4.82moles)* (2g/mole)=9.64g$

Therefore, the mass of $H_2$ is, 9.64 grams.

Write the condensed ground-state electron configurations of these transition metal ions, and state which are paramagnetic:(a) V³⁺ (b) Cd²⁺ (c) Co³⁺ (d) Ag⁺