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2 years ago

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A 6.50-g sample of copper metal at 25.0 Â°c is heated by the addition of 84.0 j of energy. the final temperature of the copper i

s __________Â°c. the specific heat capacity of copper is

1 answer:

anzhelika [568]2 years ago

3 0

The final temperature of the copper is 59.0. The specific heat capacity of copper is 0.38 j/g -k

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(a) calculate %ic of the interatomic bonds for the intermetallic compound al6mn. (b) on the basis of this result what type of in

Tatiana [17]

The answer is:

a)0.25%

b) metallic bond

The explanation:

A) The percentage of ionic character in a compound having some covalent character can be calculated by the following equation. The percent ionic character = Observed dipole moment/Calculated dipole moment assuming 100% ionic bond × 100.

-The percent ionic character is a function of the electron negativities of the ions XA and XB . The electronegativities for Al and Mn are 1.5 and 1.6, respectively :

when %IC = [1-exp(-1/4) (XB-XA)^2].100

so, %IC = [1 - exp(- 0.25)(1.6- 1.5)^2] . 100 = 0.25%

(b) Because the percent ionic character is so small (0.25%) and this intermetallic compound is composed of two metals Al and Mn, the bonding is completely metallic.

Metallic bond:

• Metallic bonding can be either weak (68 kJ/mole or 0.7 eV/atom for Hg) or strong (850 kJ/mole or 8.8 eV/atom for W)

• Metallic bonding gives rise to high electrical and thermal conductivity.

-The electrons are loosely held since each atom has several unoccupied valence orbitals; it is relatively easy for the electrons to move about. In this manner the electrons allow atoms to slide past each other.

6 0

2 years ago

After going through a guided tutorial by selecting Run Grams Demonstration, you can create your own experiment by clicking the R

Step2247 [10]

Answer: Thus 24.0 g of $SO_2$ would be needed.

Explanation:

To calculate the moles :

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

$\text{Moles of} O_2=\frac{6.00g}{32g/mol}=0.1875moles$

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(l)$

According to stoichiometry :

1 mole of $O_2$ require = 2 moles of $SO_2$

Thus 0.1875 moles of $O_2$ will require= $\frac{2}{1}\times 0.1875=0.375moles$ of $SO_2$

Mass of $SO_2=moles\times {\text {Molar mass}}=0.375moles\times 64g/mol=24.0g$

Thus 24.0 g of $SO_2$ would be needed to completely react with 6.00 g of $O_2$ such that all reactants could be consumed.

5 0

2 years ago

A student reacts 20.0 mL of 0.248 M H2SO4 with 15.0 mL of 0.195 M NaOH. Write a balanced chemical equation to show this reaction

Molodets [167]

<u>Answer:</u> The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

<u>Explanation:</u>

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

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

<u>For NaOH:</u>

Initial molarity of NaOH solution = 0.195 M

Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol$

<u>For sulfuric acid:</u>

Initial molarity of sulfuric acid solution = 0.248 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, $2.925\times 10^{-3}$ moles of KOH will react with = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sulfuric acid

As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of sulfuric acid = $(4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol$

By Stoichiometry of the reaction:

2 moles of KOH produces 1 mole of sodium sulfate

So, $2.925\times 10^{-3}$ moles of KOH will produce = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sodium sulfate

<u>For sodium sulfate:</u>

Moles of sodium sulfate = $1.462\times 10^{-3}moles$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M$

<u>For sulfuric acid:</u>

Moles of excess sulfuric acid = $3.498\times 10^{-3}mol$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M$

<u>For NaOH:</u>

Moles of NaOH remained = 0 moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of NaOH}=\frac{0}{0.050}=0M$

Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

8 0

2 years ago

If a sample of gas is at 672 °C and 3.9 atm, what is the new temperature at 12.2 atm?

Snowcat [4.5K]

In order to calculate the final temperature of the gas, we may apply Charles's law, which states that the pressure and temperature of a fixed amount of gas at constant volume are directly proportional. Mathematically:
P/T = constant
(absolute temperature is used, so T = 672 + 273 = 945 K)
Thus,
3.9 / 945 = 12.2 / T
T = 2,956 K or 2,683 °C

7 0

2 years ago

Convert 1.21 kg to grams

MissTica

Answer:

1210 grams

Explanation:

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snap- parkguy786

7 0

2 years ago

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