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3 years ago

Is The atoms of the elements have different masses but the same volume true?

Chemistry

1 answer:

stepladder [879]3 years ago

5 0

Yes, the atoms of the elements do have different masses but the same volume

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SIZIF [17.4K]

Answer: 0.9375 g

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}$ .....(1)

Molarity of $HCl$ solution = 0.75 M

Volume of $HCl$ solution = 25.0 mL = 0.025 L

Putting values in equation 1, we get:

$\text{Moles of} HCl={0.75}\times{0.025}=0.01875moles$

$CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(s)+CO_2(g)+H_2O(l)$

According to stoichiometry :

2 moles of $HCl$ require = 1 mole of $CaCO_3$

Thus 0.01875 moles of $HCl$ will require= $\frac{1}{2}\times 0.01875=0.009375moles$ of $CaCO_3$

Mass of $CaCO_3=moles\times {\text {Molar mass}}=0.009375moles\times 100g/mol=0.9375g$

Thus 0.9375 g of $CaCO_3$ is required to react with 25.0 ml of 0.75 M HCl

6 0

2 years ago

1 C8H10(l) +21/2O2(g) → 8CO2(g) + 5H2O(g), Hcomb= ? Hf for C8H10(l) = +49.0kJ/mol C8H10(l) Use the balanced combustion reaction

nikklg [1K]

Answer:

$H_{comb}=-4406kJ/mol$

Explanation:

Hello,

In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

$H_{comb}=8*\Delta _fH_{CO_2}+5\Delta _fH_{H_2O}-\Delta _fH_{C_8H_{10}}$

Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:

$H_{comb}=8*(-393.5kJ/mol)+5(-241.8kJ/mol)-49.0kJ/mol\\\\H_{comb}=-4406kJ/mol$

Best regards.

4 0

3 years ago

Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi

bija089 [108]

Answer: The value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

Explanation:

The given chemical equations follows:

Equation 1: $B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b$

Equation 2: $H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}$

The net equation follows:

$B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c$

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=K_1\times K_2$