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3 years ago

The first step of crystal formation is nucleation Is this true or false?

Chemistry

1 answer:

ankoles [38]3 years ago

6 0

True the first step is nucleation

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Ravi cow gives 2 1/4 gallon Milk each day . How much is this in quarts

EastWind [94]

Ooooh, i know this x'D.

There are 4 quarts in a gallon. So 2(Gallons)*4(Number of quarts per gallon) equals 8. Then we have 1/4th left, since there are 4 in each gallon, this 1/4th is a quart. Long explanation short......2*4=8+1=9

9 Quarts.

-Steel jelly

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3 years ago

The equation below is correctly balanced.<br><br> 6HgO 3O2 + 6Hg<br> True<br> False

Mama L [17]

The answer is TRUE because in that ecuation there are 6 atoms or moles of mercury in both sides and you also have the required moles of oxygen

5 0

4 years ago

At stp,250cm3 of gas had a mass of 0.36g.what result does this give for the molar mass of the gas

Dafna11 [192]

Answer:

32,256g/mol

Explanation:

n= V/Vm

n = 0.25/22.4

n = 0.01

0.01 = 0.36/M

M = 0.36/0.01

5 0

3 years ago

What is an example of a polyatomic ion

Pie

Answer: The hydroxide cation (OH -) and the phosphate cation (PO 4 3-) are both polyatomic ions.

Explanation:

5 0

3 years ago

Read 2 more answers

The activation energy for a reaction is 84.0kJ/mol. Addition of a catalyst lowers the activation energy by 23.0 kJ/mol, while le

Lynna [10]

Answer : The rate constant of the reaction is increased by factor, $4.93\times 10^{10}$

Solution :

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

The expression used with catalyst and without catalyst is,

$\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}$

$\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}$

where,

$K_2$ = rate of reaction with catalyst

$K_1$ = rate of reaction without catalyst

$Ea_2$ = activation energy with catalyst = 23.0 kJ/mol = 23000 J/mol

$Ea_1$ = activation energy without catalyst = 84.0 kJ/mol = 84000 J/mol

R = gas constant = 8.314 J/mol.K

T = temperature = $25^oC=273+25=298K$

Now put all the given values in this formula, we get

$\frac{K_2}{K_1}=e^{\frac{(84000-23000)J/mol}{8.314J/mol.K\times 298K}}$

$\frac{K_2}{K_1}=4.93\times 10^{10}$

Therefore, the rate constant of the reaction is increased by factor, $4.93\times 10^{10}$

8 0

3 years ago