Help!!! ASAP helppppp

How many moles of copper are present in a sample of tennantite with a mass of 2290 grams?

grin007 [14]

Answer:

18.58 moles of Cu

Solution:

Data Given:

Chemical Formula of Tennantite = Cu₁₂As₄S₁₃

Mass of Cu₁₂As₄S₁₃ = 2290 g

M.Mass of Cu₁₂As₄S₁₃ = 1479.06 g.mol⁻¹

Step 1: Calculate Moles of Cu₁₂As₄S₁₃ as,

Moles = Mass ÷ M.Mass

Putting values,

Moles = 2290 g ÷ 1479.06 g.mol⁻¹

Moles = 1.548 mol

Step 2: Calculate Moles of Cu,

As,

1 mole of Cu₁₂As₄S₁₃ contains = 12 moles of Cu

So,

1.548 mol of Cu₁₂As₄S₁₃ will contain = X moles of Cu

Solving for X,

X = (1.548 mol × 12 mol) ÷ 1 mol

X = 18.58 moles of Cu

5 0

4 years ago

Answer please bruhs and wy7uhjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjt

allsm [11]

Answer:

it's going to be A trust me

3 0

3 years ago

Read 2 more answers

My labmate discovered a new compound and wanted to try to determine what it could be. We noticed that when mixed with an acid, i

kvv77 [185]

Answer:

it went under the process known as crystalization

Explanation:

am a chemist

4 0

3 years ago

Lead(II) nitrate is added slowly to a solution that is 0.0800 M in CT ions. Calculate the concentration of Pb2+ ions (in mol/L)

charle [14.2K]

Answer : The concentration of $Pb^{2+}$ ion is 0.0375 M.

Explanation :

The balanced equilibrium reaction will be:

$Pb^{2+}+2Cl^-\rightleftharpoons PbCl_2$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Pb^{2+}][Cl^-]^2$

Now put all the given values in this expression, we get:

$2.40\times 10^{-4}=[Pb^{2+}]\times (0.0800)^2$

$[Pb^{2+}]=0.0375M$

Therefore, the concentration of $Pb^{2+}$ ion is 0.0375 M.

8 0

3 years ago

You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $200mL+200L=400mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 400 g

$T_{final}$ = final temperature of water = $27.78^oC=273+25.10=300.78K$

$T_{initial}$ = initial temperature of metal = $25.10^oC=273+27.78=298.1K$

Now put all the given values in the above formula, we get:

$q=400g\times 4.18J/g^oC\times (300.78-298.1)K$

$q=4480.96J$

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

$\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0

3 years ago