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3 years ago

10

Please help!! A lot of points!!

1 answer:

Artist 52 [7]3 years ago

7 0

Mean: the average. you have to add the values of the numbers and then divide by the amount of numbers there are. a common mistake to avoid is forgetting to divide the numbers at the end or subtracting them instead of adding.

mean: the middle number. you would first need to order the numbers from least to greatest. a common mistake to avoid is finding the middle number before ordering it from least to greatest

these two can also be commonly mistaken for one another because of the similar spelling.

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What are two lipids that store energy

timurjin [86]

Fats and oils are two lipids that store energy

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3 years ago

1.....What are the 2 major divisions of Biochemistry?

kicyunya [14]

Answer:

Organic and inorganic

Explanation:

6 0

3 years ago

A sample of an unknown liquid has a volume of 30.0 mL and a mass of 6 g. What is its density? please Show your work or explain h

Leno4ka [110]

The density of the liquid is 0.2 g/mL.

The mass of the liquid is 6 g.

The volume of the liquid is 30.0 mL.

Density = mass/volume = 6 g/30.0 mL = 0.2 g/mL

3 0

3 years ago

Calculate the ph at the equivalence point for the titration of 0.230 m methylamine (ch3nh2) with 0.230 m hcl. the kb of methylam

Vlad1618 [11]

Answer: The pH at the equivalence point for the titration will be 0.65.

Solution:

Let the concentration of $[OH^-]$ be x

Initial concentration of $[CH3NH_2]$ , c = 0.230 M

$CH_3NH_2+H_2O\rightleftharpoons CH_3NH_3^++OH^-$

at eq'm c-x x x

Expression of $K_b$ :

$K_b=\frac{[CH_3NH_3^+][+OH^-]}{[CH_3NH_2]}=\frac{x\times x}{c-x}=\frac{x^2}{c-x}$

Since ,methyl-amine is a weak base,c>>x so $c-x\approx c$ .

$K_b=\frac{x^2}{c}=5.0\times 10^{-4}=\frac{x^2}{0.230 M}$

Solving for x, we get:

$x=1.07\times 10^{-2} M$

Given, HCl with 0.230 M , it dissociates fully in water which means $[H^+]$ = 0.230 M

$[OH^-]=[H^+]$ will result in neutral solution, since $[OH^-]$

Remaining $[H^+]$ after neutralizing $[OH^-]$ ions

$[H^+]_{\text{left in solution}}=[H^+]-[OH^-]=0.230-1.07\times 10^{-2}=0.2193 M$

$pH=-log{[H^+]_{\text{left in solution}}=-log(0.2193)=0.65$

The pH at the equivalence point for the titration will be 0.65.

8 0

4 years ago

A large cylindrical tank contains 0.750 m3 of nitrogen gas at 27°C and 7.50 * 103 Pa (absolute pressure). The tank has a tight‐f

Nuetrik [128]

Answer: The answer is 68142.4 Pa

Explanation:

Given that the initial properties of the cylindrical tank are :

Volume V1= 0.750m3

Temperature T1= 27C

Pressure P1 =7.5*10^3 Pa= 7500Pa

Final properties of the tank after decrease in volume and increase in temperature :

Volume V2 =0.480m3

Temperature T2 = 157C

Pressure P2 =?

Applying the gas law equation (Charles and Boyle's laws combined)

P1V1/T1 = P2V2/T2

(7500 * 0.750)/27 =( P2 * 0.480)/157

P2 =(7500 * 0.750* 157) / (0.480 *27)

P2 = 883125/12.96

P2 = 68142.4Pa

Therefore the pressure of the cylindrical tank after decrease in volume and increase in temperature is 68142.4Pa

8 0

3 years ago