CHEMISTRY HIGH SCHOOL

1. For each of the following numbers, by how many places must the decimal point bemoved to express the number in scientific notation?
II. For each number, will the exponent be positive, negative, or zero?
0.0000089911

Answers

Answer 1
Answer: The number that you have shown the exponent would be negative and I believe it would be 6 places so 8.9911 x 10^-6

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The reaction A→B has been experimentally determined to be second order. The initial rate is 0.0100M/s at an initial concentration of A of 0.300M. Determine the initial rate at [A]=0.900M.

Answers

Answer:

0.09 M/s is the initial rate when concentration of reactant A is 0.900 M.

Explanation:

A → B

1) Initial concentration of A = [A] =0.300 M

Rate constant of the reaction , k = ?

Rate equation for the second order kinetic is given as:

R=k[A]^2

0.0100M/s=k[0.300 M]^2

k=(0.0100M/s)/([0.300 M]^2)=0.1111 M^(-1) s^(-1)

2) Rate of the reaction when initial concentration of A was 0.900M be R'.

[A] = 0.900M

R'=k[A]^2

R'=0.1111 M^(-1) s^(-1)* [0.900 M]^2

R'=0.09 M/s

The atomic number and mass number do not change during this type of radioactive decay.

Answers

In gamma decay, no change in proton number occurs, so the atom does not become a different element

Which structure is found in a plant cell but not in a animal cell is ita. chloroplast
b. cell membrane or
c. ribosome

Answers

The one on the left is an animal cell and the one on the right is a plant cell, the answer is chloroplast

Answer:

A.) Chloroplast

Explanation:

Chloroplasts are found in plant cells but not in animal cells. The chloroplast is the organelle that captures light energy and is the site where photosynthesis occurs.

An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (d = 1.114 g/mL; M = 62.07 g/mol) and water (d = 1.00 g/mL) at 20°C. The density of the mixture is 1.070 g/mL. Express the concentration of ethylene glycol as (a) volume percent 50 % v/v (b) mass percent 52.7 % w/w (c) molarity M (d) molality m (e) mole fraction

Answers

Answer :

(a) The volume percent is, 50.63 %

(b) The mass percent is, 52.69 %

(c) Molarity is, 9.087 mole/L

(d) Molality is, 17.947 mole/L

(e) Moles fraction of ethylene glycol is, 0.244

Explanation : Given,

Density of ethylene glycol = 1.114 g/mL

Molar mass of ethylene glycol = 62.07 g/mole

Density of water = 1.00 g/mL

Density of solution or mixture = 1.070 g/mL

According to the question, the mixture is made by mixing equal volumes of ethylene glycol and water.

Suppose the volume of each component in the mixture is, 1 mL

First we have to calculate the mass of ethylene glycol.

\text{Mass of ethylene glycol}=\text{Density of ethylene glycol}* \text{Volume of ethylene glycol}=1.114g/mL* 1mL=1.114g

Now we have to calculate the mass of water.

\text{Mass of water}=\text{Density of water}* \text{Volume of water}=1.00g/mL* 1mL=1.00g

Now we have to calculate the mass of solution.

Mass of solution = Mass of ethylene glycol + Mass of water

Mass of solution = 1.114 + 1.00 = 2.114 g

Now we have to calculate the volume of solution.

\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=(2.114g)/(1.070g/mL)=1.975mL

(a) Now we have to calculate the volume percent.

\text{Volume percent}=\frac{\text{Volume of ethylene glycol}}{\text{Volume of solution}}* 100=(1mL)/(1.975mL)* 100=50.63\%

(b) Now we have to calculate the mass percent.

\text{Mass percent}=\frac{\text{Mass of ethylene glycol}}{\text{Mass of solution}}* 100=(1.114g)/(2.114g)* 100=52.69\%

(c) Now we have to calculate the molarity.

\text{Molarity}=\frac{\text{Mass of ethylene glycol}* 1000}{\text{Molar mass of ethylene glycol}* \text{Volume of solution (in mL)}}

\text{Molarity}=(1.114g* 1000)/(62.07g/mole* 1.975L)=9.087mole/L

(d) Now we have to calculate the molality.

\text{Molality}=\frac{\text{Mass of ethylene glycol}* 1000}{\text{Molar mass of ethylene glycol}* \text{Mass of water (in g)}}

\text{Molality}=(1.114g* 1000)/(62.07g/mole* 1kg)=17.947mole/kg

(e) Now we have to calculate the mole fraction of ethylene glycol.

\text{Mole fraction of ethylene glycol}=\frac{\text{Moles of ethylene glycol}}{\text{Moles of ethylene glycol}+\text{Moles of water}}

\text{Moles of ethylene glycol}=\frac{\text{Mass of ethylene glycol}}{\text{Molar of ethylene glycol}}=(1.114g)/(62.07g/mole)=0.01795mole

\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar of water}}=(1g)/(18g/mole)=0.0555mole

\text{Mole fraction of ethylene glycol}=(0.01795mole)/(0.01795mole+0.0555mole)=0.244

when you check a toaster with a digital or analog video in which one of the following indicates dirty contacts on the thermostat blade

Answers

i sow it the answer if you serarch it in the box i see it the same question!

Answer:

Explanation:current higher

Than the rated current

Persamaan kimia bagi magnesium

Answers

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