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3 years ago

The pendulum has the greatest kinetic energy and no potential energy when it is _____.

Chemistry

1 answer:

Bad White [126]3 years ago

3 0

Answer:

At the top of the swing

Explanation:

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What amount of heat is produced during the combustion of hexane if the temperature of the calorimeter increases by 4.542 k?

Flauer [41]

Answer:

the answer is 85.66

Explanation:

5 0

2 years ago

Calculate the amount of heat water absorbs from a piece of hot metal using the following data: 75.0 g of cold water is placed in

AysviL [449]

Answer:

Amount of heat absorbed by water is 2604.54 J.

Explanation:

Amount of heat absorbed by water = $m_{water}\times C_{water}\times \Delta T_{water}$

where m represents mass, C represents specific heat and $\Delta T$ represents change in temperature.

Here $m_{water}=75.0$ g , $C_{water}=4.184J/(g.^{0}\textrm{C})$ and $\Delta T$ = (final temperature - initial temperature) = (29.5-21.2) $^{0}\textrm{C}$ = 8.3 $^{0}\textrm{C}$

So, amount of heat heat absorbed by water

= $(75.0g)\times (4.184\frac{J}{g.^{0}\textrm{C}})\times (8.3^{0}\textrm{C})$

= 2604.54 J

8 0

3 years ago

The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of

timama [110]

Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution = $1.83g/cm^3=1.83g/ml$

Molar mass of sulfuric acid (solute) = 98.079 g/mole

98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

Mass of sulfuric acid (solute) = 98.0 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.83g/ml}=54.64ml$

Now we have to calculate the molarity of solution.

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L$

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg$

Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

7 0

3 years ago

If 40.0 mL of a calcium nitrate solution reacts with excess potassium carbonate to yield 0.524 grams of a precipitate, what is t

tankabanditka [31]

Answer : The molarity of calcium ion on the original solution is, 0.131 M

Explanation :

The balanced chemical reaction is:

$Ca(NO_3)_2+K_2CO_3\rightarrow CaCO_3+3KNO_3$

When calcium nitrate react with potassium carbonate to give calcium carbonate as a precipitate and potassium nitrate.

First we have to calculate the moles of $CaCO_3$

$\text{Moles of }CaCO_3=\frac{\text{Mass of }CaCO_3}{\text{Molar mass of }CaCO_3}$

Given:

Mass of $CaCO_3$ = 0.524 g

Molar mass of $CaCO_3$ = 100 g/mol

$\text{Moles of }CaCO_3=\frac{0.524}{100g/mol}=0.00524mol$

Now we have to calculate the concentration of $CaCO_3$

$\text{Concentration of }CaCO_3=\frac{\text{Moles of }CaCO_3}{\text{Volume of solution}}=\frac{0.00524mol}{0.040L}=0.131M$

Now we have to calculate the concentration of calcium ion.

As, calcium carbonate dissociate to give calcium ion and carbonate ion.

$CaCO_3\rightarrow Ca^{2+}+CO_3^{2-}$

So,

Concentration of calcium ion = Concentration of $CaCO_3$ = 0.131 M

Thus, the concentration or molarity of calcium ion on the original solution is, 0.131 M

4 0

3 years ago

How many milliliters of a 0.285 M HCl solution are needed to neutralize 249 mL of a 0.0443 M Ba(OH)2 solution?

meriva

Answer:

$\large \boxed{\text{77.4 mL}}$

Explanation:

Ba(OH)₂ + 2HCl ⟶ BaCl₂ + H₂O

V/mL: 249

c/mol·L⁻¹: 0.0443 0.285

1. Calculate the moles of Ba(OH)₂

$\text{Moles of Ba(OH)$_{2}$} = \text{0.249 L Ba(OH)}_{2} \times \dfrac{\text{0.0443 mol Ba(OH)}_{2}}{\text{1 L Ba(OH)$_{2}$}} = \text{0.011 03 mol Ba(OH)}_{2}$

2. Calculate the moles of HCl

The molar ratio is 2 mol HCl:1 mol Ba(OH)₂

$\text{Moles of HCl} = \text{0.011 03 mol Ba(OH)}_{2} \times \dfrac{\text{2 mol HCl}}{\text{1 mol Ba(OH)}_{2}} = \text{0.022 06 mol HCl}$

3. Calculate the volume of HCl

$V_{\text{HCl}} = \text{0.022 06 mol HCl} \times \dfrac{\text{1 L HCl}}{\text{0.285 mol HCl}} = \text{0.0774 L HCl} = \textbf{77.4 mL HCl}\\\\\text{You must add $\large \boxed{\textbf{77.4 mL}}$ of HCl.}$

8 0

3 years ago