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3 years ago

How many moles of KMnO, are in 66.38 g of KMnO?

Chemistry

1 answer:

schepotkina [342]3 years ago

5 0

Answer:.603moles

Explanation:do given over 1 so 66.38 over 1 then multiply by 1 over 110.035344(the atomic mass of KMnO) and then you get the answear

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Suppose you have 4.0 L of helium in a flexible container (like a balloon) at a certain temperature and pressure. What will the v

AleksandrR [38]

Answer:

Volume of the helium will be 1.0L

Explanation:

To understand this question it is necessary to talk about 2 laws:

Charles's law that states: The volume of a given amount of gas held at constant pressure is directly proportional to absolute temperature.

Boyle's law that states: The volume of a given amount of gas held at constant temperature varies inversely with the applied pressure when the temperature and mass are constant.

Now, in the problem, if you have 4.0L of helium:

If the temperature is halved and the volume varies directly with temperature (Charles's law), the volume of helium will be 2.0L

Then, the pressure es doubled and as volume varies inversely with pressure (Boyle's law), the volume of helium will be halved, being 1.0L.

Thus, volume of the helium will be 1.0L

5 0

3 years ago

Read 2 more answers

Word equation for Na2O+H2O → NaOH

Yuri [45]

Answer:

Sodium oxide + Water - - - > Sodium Hydroxide.

8 0

3 years ago

Help me please............…

Simora [160]

Answer:I think this is right: 1. 6 2. 12 3. 4 4. In the middle 5. Carbon 1. The one on the right 2a. Beryllium 2b. Aluminium

Explanation:The protons are the circles with plus signs and the black ones are nutrons. I looked on my periodic table for the rest.

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4 years ago

73.5 mol of P4O10 contains how many moles of P ? moles of P:

Sholpan [36]

Answer:

294 moles of P

Explanation:

For every 1 mol of P4O10 contains 4 mol of P

so;

73.5 mol P4O10 × 4 mol P

1 mol P4O10

= 73.5 × 4

= 294 moles of P

7 0

2 years ago

The ksp of manganese(ii) carbonate, mnco3, is 2.42 × 10-11. calculate the solubility of this compound in g/l.

Vikki [24]

Answer : The solubility of this compound in g/L is $565.414\times 10^{-6}g/L$ .

Solution : Given,

$K_{sp}=2.42\times 10^{-11}$

Molar mass of $MnCO_3$ = 114.945g/mole

The balanced equilibrium reaction is,

$MnCO_3\rightleftharpoons Mn^{2+}+CO^{2-}_3$

At equilibrium s s

The expression for solubility constant is,

$K_{sp}=[Mn^{2+}][CO^{2-}_3]$

Now put the given values in this expression, we get

$2.42\times 10^{-11}=(s)(s)\\2.42\times 10^{-11}=s^2\\s=0.4919\times 10^{-5}=4.919\times 10^{-6}moles/L$

The value of 's' is the molar concentration of manganese ion and carbonate ion.

Now we have to calculate the solubility in terms of g/L multiplying by the Molar mass of the given compound.

$s=4.919\times 10^{-6}moles/L\times 114.945g/mole=565.414\times 10^{-6}g/L$

Therefore, the solubility of this compound in g/L is $565.414\times 10^{-6}g/L$ .

8 0

3 years ago