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3 years ago

How many molecules are in 4.0 mol h20

Chemistry

1 answer:

Leona [35]3 years ago

7 0

4 moles of water will have 4 number of water molecules

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Electrons is an excited state are more likely to enter into chemical reactions.

Alex17521 [72]

A. True would be the best answer

3 0

3 years ago

What is the molecular formula of a compound with the empirical formula

ipn [44]

Answer: $C_2H_2O_2$

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

The empirical formula is $CHO$

The empirical weight of $CHO$ = 1(12)+1(1)+1(16)= 29 g.

The molecular weight = 60 g/mole

Now we have to calculate the molecular formula:

$n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{60}{29}=2$

The molecular formula will be= $2\times CHO=C_2H_2O_2$

Thus molecular formula will be $C_2H_2O_2$

5 0

3 years ago

How is kinetic energy of particles and temperature related

Rufina [12.5K]

Answer:

The higher the temperature, the faster the particles move, the lower the temperature, the slower.

4 0

3 years ago

What is the mass of 3.01x1023 atoms of iron(atomic mass of fe=56)

nlexa [21]

Answer:

N = n× l

N = number of entities

n= moles

l = Avogadro's constant = 6.023 × 10^23

3.01 × 10^ 23 = n * 6.023 × 10^23

n = 3.01 × 10^23/6.023 × 10^23

n= 0.5moles

Molar mass = mass/ number of moles

Molar mass = 56

mass = 56 × 0.5

= 28g

Hope this helps.

4 0

3 years ago

Consider the reaction: P(s) + 5/2 Cl2(g)PCl5(g) Write the equilibrium constant for this reaction in terms of the equilibrium con

Pani-rosa [81]

Answer: The equilibrium constant for the overall reaction is $K_a\times K_b$

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.

a) $P(s)+\frac{3}{2}Cl_2(g)\rightarrow PCl_3(g)$

$K_a=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}$

b) $PCl_3(g)+Cl_2(g)\rightarrow PCl_5(g)$

$K_b=\frac{[PCl_5]}{[Cl_2]\times [PCl_3]}$

For overall reaction on adding a and b we get c

c) $P(s)+\frac{5}{2}Cl_2(g)\rightarrow PCl_5(g)$

$K_c=\frac{[PCl_5]}{[Cl_2]^\frac{5}{2}}$

$K_c=K_a\times K_b=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}\times \frac{[PCl_5]}{[Cl_2]\times [PCl_3]}$

The equilibrium constant for the overall reaction is $K_a\times K_b$

4 0

3 years ago