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3 years ago

Calculate the molecular weight of chalk.write any two advantages of mandeleev's periodic table?

1 answer:

5 0

Part (a):
Chalk has the chemical formula: CaCo₃
From the periodic table:
mass of Ca = 40 grams
mass of Co = 58.9 grams
Therefore:
molecular weight of chalk = 40 + 3(58.9) = 216.7 grams

Part (b):
Advantages of Mandeleev's periodic table:
1- Could predict the properties of elements easily based on its position in the periodic table
2- Elements were arranged systematically in order of increasing atomic mass which helped in the study and classification of elements
3- Even after the discovery of noble gases, no major changes needed to be done to the periodic table. Noble gases were inserted easily
4- He left room for other undiscovered elements based on his systematic approach in writing the table

Hope this helps :)

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2 HI(g) ⇄ H2(g) + I2(g) Kc = 0.0156 at 400ºC 0.550 moles of HI are placed in a 2.00 L container and the system is allowed to rea

Alex_Xolod [135]

Answer: The concentration of hydrogen gas at equilibrium is 0.0275 M

Explanation:

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}$

Moles of HI = 0.550 moles

Volume of container = 2.00 L

$\text{Initial concentration of HI}=\frac{0.550}{2}=0.275M$

For the given chemical equation:

$2HI(g)\rightleftharpoons H_2(g)+I_2(g)$

Initial: 0.275

At eqllm: 0.275-2x x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[H_2][I_2]}{[HI]^2}$

We are given:

$K_c=0.0156$

Putting values in above expression, we get:

$0.0156=\frac{x\times x}{(0.275-2x)^2}\\\\x=-0.0458,0.0275$

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.0275 M

Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M

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3 years ago

How was your great day

Alexandra [31]

Lolol its going good wbu

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2 years ago

(c) O2 gas is transferred from a 3 L vessel containing oxygen at 4 atm to an evacuated 20 L vessel at a constant temperature of

Fudgin [204]

Answer:

After the transfer the pressure inside the 20 L vessel is 0.6 atm.

Explanation:

Considering O2 as an ideal gas, it is at an initial state (1) with V1 = 3L and P1 = 4 atm. And a final state (2) with V2 = 20L. The temperature remain constant at all the process, thus here applies the Boyle-Mariotte law. This law establishes that at a constant temperature an ideal gas the relationship between pressure and volume remain constant at all time:

$P x V = k$

Therefore, for this problem the step by step explanation is:

$P_{1} xV_{1} = P_{2} xV_{2}$

Clearing P2 and replacing

$P_{2}= \frac{P_{1} xV_{1}}{V_{2} } = \frac{4atmx3L}{20L} = 0.6atm$

7 0

3 years ago

Please review the attachment

astra-53 [7]

Answer: The correct answer is -297 kJ.

Explanation:

To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:

2SO3 —> O2 + 2SO2 (196 kJ)

Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.

Now, both equations have double the coefficients that we would like (for example, there is 2S in the second equation when we need only S). This means we should multiply each equation (and their enthalpy changes) by 1/2. This gives us:

SO3 —>1/2O2 + SO2 (98 kJ)

S + 3/2O2 —> SO3 (-395 kJ)

Now, we add the two equations together. Notice that the SO3 in the reactants in the first equation and the SO3 in the products of the second equation cancel. Also note that O2 is present on both sides of the equation, so we must subtract 3/2 - 1/2, giving us a net 1O2 on the left side of the equation.

S + O2 —> SO2

Now, we must add the enthalpies together to get our final answer.

-395 kJ + 98 kJ = -297 kJ

Hope this helps!

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3 years ago

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Dominik [7]

Answer:

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3 years ago

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