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2 years ago

what volume of hydrogen gas is evolved from a reaction between 0.52 g of Na and water? This gas is collected at 20 C and 745mmHg

1 answer:

6 0

The volume of hydrogen gas that evolved is calculated as follows
by use of ideal gas equation

that is PV = nRT
P=745 mm hg
V= ?
R(gas constant)= 62.36 L.mm hg/mol.k
T= 20 + 273 = 293 k
n=number of moles which is calculated as follows
find the moles of Na used
= 0.52/23=0.023 moles

write the reacting equation
2Na +2H2O =2NaOH +H2
by use of reacting ratio between Na : H2 which is 2:1 therefore the mole of H2 = 0.023/2 =0.0115 moles

by making the volume the subject of the formula
v=nRT/P
V= (0.0115 x 62.36 x 293) / 745 = 0.283 L

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2 years ago

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The average atomic weight of copper, which has two naturally occurring isotopes, is 63.5. One of the isotopes has an atomic weig

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Answer: The atomic weight of the second isotope is 64.81 amu.

Explanation:

Average atomic mass of an element is defined as the sum of atomic masses of each isotope each multiplied by their natural fractional abundance

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

We are given:

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Percentage abundance of isotope 1 = 69.1 %

Fractional abundance of isotope 1 = 0.691

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Percentage abundance of isotope 2 = 30.9%

Fractional abundance of isotope 2 = 0.309

Average atomic mass of copper = 63.5 amu

Putting values in equation 1, we get:

$\text{Average atomic mass of copper}=[(62.9\times 0.691)+(x\times 0.309)]$

$x=64.81amu$

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2 years ago

What happens when the calcium burns in the oxygen. write the chemical reaction

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2 years ago

Which is an unlikely isotope of Al? Al-26 Al-25 Al-27 Al-72

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2 years ago

If 500.0 ml of 0.10 m ca2+ is mixed with 500.0 ml of 0.10 m so42−, what mass of calcium sulfate will precipitate? ksp for caso4

attashe74 [19]

Answer : The mass of calcium sulfate precipitate will be, 6.12 grams

Solution :

First we have to calculate the moles of $Ca^{2+}$ and $SO_4^{2-}$ .

$\text{Moles of }Ca^{2+}=\text{Molarity of }Ca^{2+}\times \text{Volume of }Ca^{2+}=0.10mole/L\times 0.5L=0.05\text{ moles}$

$\text{Moles of }SO_4^{2-}=\text{Molarity of }SO_4^{2-}\times \text{Volume of }SO_4^{2-}=0.10mole/L\times 0.5L=0.05\text{ moles}$

As, 0.05 moles of $Ca^{2+}$ is mixed with 0.05 moles of $SO_4^{2-}$ , it gives 0.05 moles of calcium sulfate.

Now we have to calculate the solubility of calcium sulfate.

The balanced equilibrium reaction will be,

$CaSO_4\rightleftharpoons Ca^{2+}+SO_4^{2-}$

The expression for solubility constant for this reaction will be,

$K_{sp}=(s)\times (s)$

$K_{sp}=(s)^2$

Now put the value of $K_{sp}$ in this expression, we get the solubility of calcium sulfate.

$2.40\times 10^{-5}=(s)^2$

$s=4.89\times 10^{-3}M$

Now we have to calculate the moles of dissolved calcium sulfate in one liter solution.

$\text{Moles of }CaSO_4=\text{Molarity of }CaSO_4\times \text{Volume of }CaSO_4=4.89\times 10^{-3}mole/L\times 1L=4.89\times 10^{-3}\text{ moles}$

Now we have to calculate the moles of calcium sulfate that precipitated.

$\text{Moles of }CaSO_4\text{ precipitated}=\text{Moles of }CaSO_4\text{ present}-\text{Moles of }CaSO_4\text{ dissolved}$

$\text{Moles of }CaSO_4\text{ precipitated}=0.05-4.89\times 10^{-3}=0.045\text{ moles}$

Now we have to calculate the mass of calcium sulfate that precipitated.

$\text{Mass of }CaSO_4\text{ precipitated}=\text{Moles of }CaSO_4\text{ precipitated}\times \text{Molar mass of }CaSO_4$

$\text{Mass of }CaSO_4\text{ precipitated}=0.045moles\times 136g/mole=6.12g$

Therefore, the mass of calcium sulfate precipitate will be, 6.12 grams

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2 years ago

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