What is 1.5 x 10^3 in standard notation

Lakesha gave three tenths of her cookies to Bailey and five tenths of her cookies to Helen. What fraction of her cookies did Lak

Mashutka [201]

Answer:

a . eight tenths of her cookies

Explanation:

Let the total number of Lakesha's cookies be represented by x.

So that;

She gave three tenths to Bailey = $\frac{3}{10}$ of x

= $\frac{3x}{10}$

She gave five tenths to Helen = $\frac{5}{10}$ of x

= $\frac{5x}{10}$

Fraction of Lakesha's cookies given away = $\frac{3x}{10}$ + $\frac{5x}{10}$

= $\frac{3x+ 5x}{10}$

= $\frac{8x}{10}$

Thus, the fraction of cookies given away by Lakesha is $\frac{8}{10}$ .

6 0

2 years ago

A 0.4021-g sample of a purified organic acid was dissolved in water and titrated potentiometrically. A plot of the data revealed

Archy [21]

Answer:

The molar mass of the acid is 167.5 g/mol

Explanation:

A 0.4021-g sample of a purified organic acid was dissolved in water and titrated potentiometrically. A plot of the data revealed a single end point after 19.31 mL of 0.1243 M NaOH had been introduced. Calculate the molecular mass of the acid.

Step 1: Data given

Mass of the sample of a purified organic acid = 0.4021 grams

Molarity = 0.1243 M

Volume needed to reach the end point = 19.1 mL = 0.01931 L

Step 2: Calculate the number of moles NaOH

Moles NaOH = molarity NaOH * volume

Moles NaOH = 0.1243 M * 0.01931 L

Moles NaOH = 0.00240 moles

Step 3: Calculate moles of the acid

We'll need 0.00240 moles of acid to neutralize 0.00240 moles of NaOH ( it's a single end point)

Moles acid = 0.00240 moles

Step 4: Calculate molar mass of the acid

Molar mass = mass / moles

Molar mass = 0.4021 grams / 0.00240 moles

Molar mass = 167.5 g/mol

The molar mass of the acid is 167.5 g/mol

6 0

3 years ago

Calculate the molar solubility of CaF2 in a 0.25 m solution of NaF(aq).

Kipish [7]

Answer:

6.4 × 10^-10 M

Explanation:

The molar solubility of the ions in a compound can be calculated from the Ksp (solubility constant).

CaF2 will dissociate as follows:

CaF2 ⇌Ca2+ + 2F-

1 mole of Calcium ion (x)

2 moles of fluorine ion (2x)

NaF will also dissociate as follows:

NaF ⇌ Na+ + F-

Where Na+ = 0.25M

F- = 0.25M

The total concentration of fluoride ion in the solution is (2x + 0.25M), however, due to common ion effect i.e. 2x<0.25, 2x can be neglected. This means that concentration of fluoride ion will be 0.25M

Ksp = {Ca2+}{F-}^2

Ksp = {x}{0.25}^2

4.0 × 10^-11 = 0.25^2 × x

4.0 × 10^-11 = 0.0625x

x = 4.0 × 10^-11 ÷ 6.25 × 10^-2

x = 4/6.25 × 10^ (-11+2)

x = 0.64 × 10^-9

x = 6.4 × 10^-10

Therefore, the molar solubility of CaF2 in NaF solution is 6.4 × 10^-10M

8 0

3 years ago

Which lab safety rule do you think is the most important to follow in chemistry lab?

Flauer [41]

Answer:

Practice good personal hygiene. Wash your hands after removing gloves, before leaving the laboratory, and after handling a potentially hazardous material. While working in the laboratory, wear personal protective equipment - eye protection, gloves, laboratory coat - as directed by your supervisor.

Explanation:

7 0

3 years ago

In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691

andre [41]

Explanation:

Let us assume that the value of $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s$