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GarryVolchara [31]
3 years ago
9

How many moles are in 0.0073 kg of tantalum?

Chemistry
1 answer:
Citrus2011 [14]3 years ago
4 0
Tantalum is the 73rd element in the periodic table with an atomic mass equal to 180.95 g/mol. To determine the number of moles present in the given mass of tantalum above, we simply divide the mass by the atomic mass. 
                   number of moles = (0.0073 kg)(1000 g/ 1kg) ÷ (180.95 g/mol)
                   number of moles = 0.0403 moles
Therefore, there is approximately 0.0403 moles of tantalum in 0.0073 kg. 
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Consider the reaction pathway graph below.
Alex17521 [72]
Hello!

The reaction that the graph represents is A. Exothermic because Hrxn=-167 kJ

To calculate Hrxn we apply the following equation:

Hrxn=Hproducts-Hreagents=-625kJ-(-458kJ)=-167kJ

Looking at the graph, and at the result of the calculations, we can see that the enthalpy of the products is lower than the enthalpy of the reagents, because the sign is negative. That means that the reaction releases energy in the form of heat and that the reaction is exothermic.

Have a nice day!
6 0
4 years ago
Help!?
wolverine [178]
I think to measure the age of the rock 
6 0
3 years ago
If 100.0g of nitrogen is reacted with 100.0g of hydrogen, what is the excess reactant? What is the limiting reactant? Show your
Drupady [299]

N₂ : limiting reactant

H₂ : excess reactant

<h3>Further explanation</h3>

Given

mass of N₂ = 100 g

mass of H₂ = 100 g

Required

Limiting reactant

Excess reactant

Solution

Reaction

<em>N₂+3H₂⇒2NH₃</em>

mol N₂(MW=28 g/mol) :

\tt mol=\dfrac{mass}{MW}=\dfrac{100}{28}=3.571

mol H₂(MW= 2 g/mol) :

\tt mol=\dfrac{100}{2}=50

A method that can be used to find limiting reactants is to divide the number of moles of known substances by their respective coefficients, and small or exhausted reactans become a limiting reactants

From the equation, mol ratio N₂ : H₂ = 1 : 3, so :

\tt \dfrac{3.571}{1}\div \dfrac{50}{3}=3.571\div 16.6

N₂ becomes a limiting reactant (smaller ratio) and H₂ is the excess reactant

5 0
3 years ago
What volume of 7.25 mol/L stock solution in needed to make 3.84 L of 8.50 mol/L solution?
Serjik [45]

Answer:

4.50 L

Explanation:

First we <u>calculate how many moles are there in 3.84 L of a 8.50 mol/L solution</u>:

  • 3.84 L * 8.50 mol/L = 32.64 mol

Now, keeping in mind that

  • Concentration = Mol / Volume

we can calculate the volume of a 7.25 mol/L solution that would contain 32.64 moles:

  • Volume = Mol / Concentration
  • Volume = 32.64 mol ÷ 7.25 mol/L
  • Volume = 4.50 L

So we could take 4.50 L of the 7.25 mol/L solution and evaporate the solvent until only 3.84 L remain.

7 0
3 years ago
If decomposition of 1 mole of ethylene with a gas density of 0.215g/mL starts at 29 degree celcius and the temperature increases
stellarik [79]
We calculate first the initial pressure of the gas. 
                         P = nRT/V
                            = (0.215 g/mL)(1 mol/16 g)(0.0821 L.atm/mol K)(302K)
                             = 0.333 atm
Using the relationship between the pressure and temperature,
                       P1/T1 = P2/T2
Substituting the known values,
                        (0.333 atm) / (29 + 273) = P2/(950)
                                 P2 = 1.05 atm
7 0
4 years ago
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