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Elan Coil [88]
3 years ago
15

Define surface are tension of liquid​

Chemistry
1 answer:
rewona [7]3 years ago
7 0

The hydrological cycle refers to the circulation of water within the earth's hydrosphere in different from I. e. the liquid, solid and the gaseous forms.

You might be interested in
.800 moles of Br is how many grams
lozanna [386]

Answer:

79.904 grams.

Explanation:

The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles Bromine, or 79.904 grams.

8 0
3 years ago
1. How many grams are there in 7.5 x 1023 molecules of H2SO4?
Karo-lina-s [1.5K]

Answer:

I cannot give you all the answer but I can help you to solve those.

Explanation:

The first question:

How many grams are there in 7.5×10^{23} molecules of H_{2} SO_{4}?

So we need to find the molecular mass first, use your periodic table,

And then we can find out 2+32+16×4=98 g/mol

Then, we need to find how many moles, by using Avogadro's constant:

Avogadro's constant: 1 mole = 6.02×10^{23}

∴\frac{7.5*10^{23} }{6.02*10^{23}}=1.25 mol(2d.p.)

Lastly, find the grams using the formula M= \frac{m}{n}

m=Mn

m=1.25*98

m=122.5g

-------------------------------------------------------------------------------------------------------------

In conclusion, use those formula to help you:

M= \frac{m}{n} (which M = molecular mass(atomic mass) m=mass of the substance and n = moles)

Avogadro's constant: \frac {molecular mass} {6.02*10^{23}} = moles

5 0
3 years ago
What is the work function of gold metal in kJ/mol if light with λ = 234 nm is necessary to eject electrons?
Tju [1.3M]

Answer

512kj/mole

Explanation:

What is the work function of gold metal in kJ/mol if light with λ = 234 nm is necessary to eject electrons?

The energy can be calculated using below expresion;

E = hc/λ

Where h= planks constant= 6.626 x 10^-34

c= speed of light= 3 x 10^

λ= wavelength

hc= (6.626 x 10^-34 x 3 x 10^ 8)

=

8 0
3 years ago
How many lone pairs are around the central atom in the Lewis formula for the
barxatty [35]

Answer:

The answer to your question is None, sulfur share of its electrons

Explanation:

Just remember:

Sulfur, S, has 6 electrons in its outermost shell

Hydrogen, H, has 1 electron in its outermost shell

Oxygen, O, has 6 electrons in its outermost shell

See the picture below

The electrons of sulfur are in blue

The electrons of oxygen are in red

The electron in hydrogen is in yellow

Sulfur is the central atom and it shares all its electrons with the oxygen.

                     

8 0
4 years ago
Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to
zhenek [66]

Answer:

<u>D. It will decrease by a factor of 4</u>

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, <u><em>doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,</em></u>

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}

<u>r=[A]^{1}[B]^{2}.............(3)</u>

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' =

<u>r' = 1/4 r</u>

7 0
3 years ago
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