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4 years ago

Convert 3.0x1p^24 molecules H2O=________ mol H2O

Chemistry

1 answer:

lbvjy [14]4 years ago

5 0

Answer:

3.0x1p^24 molecules H2O=____5____ mol H2O

Explanation:

We une the number of Avogadro:

6, 02 x 10 ^23 molecules------- 1 mol

3, 01 x 10 ^24 molecules----x=

x= (3, 01 x 10 ^24 moleculesx 1 mol)/6, 02 x 10 ^23 molecules)

x= 5 mol

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4 years ago

A 27.5 −g aluminum block is warmed to 65.9 ∘C and plunged into an insulated beaker containing 55.5 g water initially at 22.1 ∘C.

Wittaler [7]

Answer:

The final temperature of aluminium ≈ 26.32 °C

Explanation:

Step 1: explain the problem

A 27.5 −g aluminum block is warmed to 65.9 ∘C and plunged into an insulated beaker containing 55.5 g water initially at 22.1 ∘C.

Step 2: Data given

We will use the formule : Q = mcΔT

with Q = heat transfer ( J)

with m = mass of the substance (g)

with c = specific heat ( J/g °C)

with ΔT = change in temperature ( in °C or K)

mass of aluminium = 27.5g

mass of water = 55.5g

specific heat of aluminium = 0.900J/g °C

specific heat of water = 4.186 J/g °C

initial temperature of aluminium T1= 65.9 °C

initial temperature of water T1 = 22.1 °C

final temperature of water and aluminium = TO BE DETERMINED

Step 3: Calculate the initial temperature

To find the final temperature, we have to use the following formule:

-(Mass of aluminium) * (caluminium)*(ΔT)) = (Mass of water) *(cwater)*(ΔT)

-27.5g (0.900)(T2 - 65.9) = 55.5g (4.184j/g °C) (T2- 22.1)

-24.75*(T2-65.9) = 232.212 *(T2-22.1)

-24.75T2 + 1631.025 = 232.212T2 -5131,8852

-256.962 T2 = -6762.9102

T2 = 26.32 °C

The final temperature of aluminium ≈ 26.32 °C

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3 years ago

Describe the appearance of the compound fes

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FeS known as iron sulphide is formed by the following reaction:
Fe + S ................> FeS

Iron sulphide has a molar mass of 87.910 grams. It has a grey color and can either be in the form of a powder of in the form of lumps.
It is insoluble in water and it reacts with acids.
Its density is 4.84 g/cm^3 and it has a melting point of 1194 degrees celcius.

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3 years ago

A reaction A(aq)+B(aq)↽−−⇀C(aq) has a standard free‑energy change of −4.20 kJ/mol at 25 °C. What are the concentrations of A, B,

Dafna1 [17]

Answer : The concentration of $A,B\text{ and }C$ at equilibrium are 0.132 M, 0.232 M and 0.168 M respectively.

Explanation :

The given chemical reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

First we have to calculate the equilibrium constant for the reaction.

The relation between the equilibrium constant and standard free‑energy is:

$\Delta G^o=-RT \ln k$

where,

$\Delta G^o$ = standard free‑energy change = -4.20 kJ/mole

R = universal gas constant = 8.314 J/mole.K

k = equilibrium constant = ?

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

Now put all the given values in the above relation, we get:

$-4.20kJ/mole=-(8.314J/mole.K)\times (298K) \ln k$

$k=5.45$

Now we have to calculate the concentrations of A, B, and C at equilibrium.

The given equilibrium reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

Initially 0.30 0.40 0

At equilibrium (0.30-x) (0.40-x) x

The expression of equilibrium constant will be,

$k=\frac{[C]}{[A][B]}$

$5.45=\frac{x}{(0.30-x)\times (0.40-x)}$

By solving the term x, we get

$x=0.168\text{ and }0.716$

From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.

The value of x will be, 0.168 M

The concentration of $A$ at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M

The concentration of $B$ at equilibrium = (0.40-x) = 0.40 - 0.168 = 0.232 M

The concentration of $C$ at equilibrium = x = 0.168 M

3 0

4 years ago

Convert 3.0x1p^24 molecules H2O=________ mol H2O​

Convert 3.0x1p^24 molecules H2O=________ mol H2O