Help SOS Quick Anyone know this?

A 10kg bowling ball would require what force to accelerate down an alleyway at a rate of 3m/s2?

mixas84 [53]

Ok so this all boils down to Newton's second law

Force = mass X acceleration

so the force = 10 X 3 = 30N

8 0

3 years ago

Camphor, a white solid with a pleasant odor, is extracted from the roots, branches, and trunk of the camphor tree. Assume you di

katrin [286]

Answer: The molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

Explanation:

Calculating the molarity of solution:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of camphor = 70.0 g

Molar mass of camphor = 152.2 g/mol

Volume of solution = 575 mL

Putting values in above equation, we get:

$\text{Molarity of camphor}=\frac{70\times 1000}{152.2\times 575}\\\\\text{Molarity of camphor}=0.799M$

Calculating the molarity of solution:

To calculate the mass of ethanol, we use the equation:

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

Density of ethanol = 0.785 g/mL

Volume of ethanol = 575 mL

Putting values in above equation, we get:

$0.785g/mL=\frac{\text{Mass of ethanol}}{575mL}\\\\\text{Mass of ethanol}=(0.785g/mL\times 575mL)=451.38g$

To calculate the molality of solution, we use the equation:

$\text{Molality of solution}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

$m_{solute}$ = Given mass of solute (camphor) = 70 g

$M_{solute}$ = Molar mass of solute (camphor) = 152.2 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 451.38 g

Putting values in above equation, we get:

$\text{Molality of camphor}=\frac{70\times 1000}{152.2\times 451.38}\\\\\text{Molality of camphor}=1.02m$

Calculating the mole fraction of camphor:

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

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

For camphor:

Given mass of camphor = 70 g

Molar mass of camphor = 152.2 g/mol

Putting values in equation 1, we get:

$\text{Moles of camphor}=\frac{70g}{152.2g/mol}=0.459mol$

For ethanol:

Given mass of ethanol = 451.38 g

Molar mass of ethanol = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of ethanol}=\frac{451.38g}{46g/mol}=9.813mol$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

Moles of camphor = 0.459 moles

Total moles = [0.459 + 9.813] = 10.272 moles

Putting values in above equation, we get:

$\chi_{(camphor)}=\frac{0.459}{10.272}=0.045$ \

Calculating the mass percent of camphor:

To calculate the mass percentage of camphor in solution, we use the equation:

$\text{Mass percent of camphor}=\frac{\text{Mass of camphor}}{\text{Mass of solution}}\times 100$

Mass of camphor = 70 g

Mass of solution = [70 + 451.38] = 521.38 g

Putting values in above equation, we get:

$\text{Mass percent of camphor}=\frac{70g}{521.38g}\times 100=13.43\%$

Hence, the molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

3 0

3 years ago

Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH. pH= 8.74, pH= 11.38, pH= 2.81

Gnom [1K]

Answer:

Explanation:

Given parameters;

pH = 8.74

pH = 11.38

pH = 2.81

Unknown:

concentration of hydrogen ion and hydroxyl ion for each solution = ?

Solution

The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.

It is graduated from 1 to 14

pH = -log[H₃O⁺]

pOH = -log[OH⁻]

pH + pOH = 14

Now let us solve;

pH = 8.74

since pH = -log[H₃O⁺]

8.74 = -log[H₃O⁺]

[H₃O⁺] = 10⁻ $^{8.74}$

[H₃O⁺] = 1.82 x 10⁻⁹mol dm³

pH + pOH = 14

pOH = 14 - 8.74

pOH = 5.26

pOH = -log[OH⁻]

5.26 = -log[OH⁻]

[OH⁻] = 10 $^{-5.26}$

[OH⁻] = 5.5 x 10⁻⁶mol dm³

2. pH = 11.38

since pH = -log[H₃O⁺]

11.38 = -log[H₃O⁺]

[H₃O⁺] = 10⁻ $^{11.38}$

[H₃O⁺] = 4.17 x 10⁻¹² mol dm³

pH + pOH = 14

pOH = 14 - 11.38

pOH = 2.62

pOH = -log[OH⁻]

2.62 = -log[OH⁻]

[OH⁻] = 10 $^{-2.62}$

[OH⁻] =2.4 x 10⁻³mol dm³

3. pH = 2.81

since pH = -log[H₃O⁺]

2.81 = -log[H₃O⁺]

[H₃O⁺] = 10⁻ $^{2.81}$

[H₃O⁺] = 1.55 x 10⁻³ mol dm³

pH + pOH = 14

pOH = 14 - 2.81

pOH = 11.19

pOH = -log[OH⁻]

11.19 = -log[OH⁻]

[OH⁻] = 10 $^{-11.19}$

[OH⁻] =6.46 x 10⁻¹²mol dm³

5 0

3 years ago

Who made the first periodic table

ale4655 [162]

The other dude is rude wrong

The correct answer is Dmitri Mendeleev

8 0

2 years ago

If a dog has a mass of 21.5kg, what is his mass in the following units?

PolarNik [594]

Answer: 21.5kg in grams is 21,500 grams

21.5kg in mg is 215,000,00

Explanation:

5 0

3 years ago