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3 years ago

Nvermind i messed u. thats not what i meant to outtb ycgdgc8ydsaox agdydac bc7w

Chemistry

1 answer:

jasenka [17]3 years ago

5 0

Answer:

same

Explanation:

yes, I agree 100% and you are correct *applaud*

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A 1 mL pipette delivers a measured volume of .98 mL (calculate the percentage error)

Natalija [7]

Answer:

Explanation:

.98 is 98% of one and therefore they are missing 2%

4 0

2 years ago

Read 2 more answers

Soap is made of molecules with atoms that share electrons equally at one end and atoms that have a slight charge at the other en

kicyunya [14]

Which could be soluble in soap?

Answer: Out of all the options presented above the one that represents which substance is soluble in soap is answer choice C) both because soap is part polar and part nonpolar.

I hope it helps, Regards.

6 0

2 years ago

Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to

zhenek [66]

Answer:

D. It will decrease by a factor of 4

Explanation:

According to the question , the equation follows :

$A+B\rightarrow C+D$

Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.

$Rate\ \alpha [A]^{a}[B]^{b}$

$r=[A]^{a}[B]^{b}$ .................(1)

STEP": First, find out the power "a" and "b"

a+b = 3 (because it is given that the reaction follow 3rd order-kinetics)

According to question, doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,

r' = 2r if [A'] = 2[A]

Here [B] is uneffected means [B']=[B]

hence new rate =

$r'=[A']^{a}[B']^{b}$

Put the value of [A'] , [B'] and r' in the above equation:

$2r=[2A]^{a}[B]^{b}$ ...........(2)

Divide equation (1) by (2) we , get

$\frac{2r}{r}=\frac{[2A]^{2}[B]^{b}}{[A]^{a}[B]^{b}}$

$2= 2(\frac{A}{A})^{a}\times (\frac{B}{B})^{b}$

Here A and A cancel each other

B and B cancel each other

We get,

$2= 2^{a}\times 1^{b}$

1^b = 1 ( power of 1 = 1)

$2= 2^{a}$

This is possible only when a = 1

We know that : a + b = 3

1 + b = 3

b =3 -1 = 2

b = 2

Hence the rate law becomes :

$r=[A]^{a}[B]^{b}$

$r=[A]^{1}[B]^{2}$ .............(3)

Look in the question now, it is asked to calculate the concentration of [B],if cut in half

Hence

[B']=1/2[B]

Insert the value of [B'] in equation (3)

$r'=[A]^{1}[B']^{2}$

$r'=[A]^{1}(\frac{1}{2}[B])^{2}$

$r'=\frac{1}{4}[A]^{1}[B]^{2}$ ............(a)

But

$r=[A]^{a}[B]^{b}$ ..............(b)

Compare equation (a) and (b) , we get

new rate r' =

r' = 1/4 r

7 0

2 years ago

A given sample of a gas has a volume of 3 liters at a

ycow [4]

Answer:

B) 12

Explanation:

Given parameters:

Initial volume = 3L

Initial pressure = 4atm

Final pressure = 6atm

Unknown:

Final volume = ?

Solution:

To solve this problem, we apply Boyle's law which states that "the volume of a fixed mass of a gas varies inversely as the pressure changes if the temperature is constant".

P₁V₁ = P₂V₂

P and V are pressure and temperature values

1 and 2 are initial and final states.

PV product = 3 x 4 = 12

7 0

3 years ago

Please help me with this! I'd appreciate it. Thanks so much! :D

Setler [38]

Answer:

It is the second option.

Explanation:

The conversion factor is:

I kPa = 7.50062 Torr.

3 0

3 years ago