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Darina [25.2K]
2 years ago
8

How many grams are in 2.12 x 10-2 moles of calcium phosphate, Ca3(PO3)2?

Chemistry
1 answer:
Vladimir79 [104]2 years ago
3 0

Answer:19.2

Explanation: if see you have to times it and show it with times

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Can anyone check my work and see if it is correct? If not, may someone help me?
shusha [124]
It looks all correct to me, great job!
6 0
3 years ago
6.0 g of copper was heated from 20 degree c to 90 degree c . How much energy was used to heat cu?
Darina [25.2K]
Copper heat capacity would be <span>0.385J/C*gram which means it needs 0.385 Joule of energy to increase 1 gram of copper temperature by 1 Celcius. The calculation would be:
energy= heat capacity *mass * temperature difference
energy= </span>0.385J/C*gram * 6g * (90-20)
<span>energy= 161.7J
 </span>
4 0
3 years ago
Calculate the standard enthalpy of formation of NOCl(g) at 25 ºC, knowing that the standard enthalpy of formation of NO(g) at th
stepan [7]

Answer:

The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol

Explanation:

The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:

Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants

In this case, you have:  2 NOCl(g) → 2 NO(g) + Cl₂(g)

So, ΔH=2*H_{NO} +H_{Cl_{2} }-2*H_{NOCl}

Knowing:

  • ΔH= 75.5 kJ/mol
  • H_{NO}= 90.25 kJ/mol
  • H_{Cl_{2} }= 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound  the chlorine Cl₂)
  • H_{NOCl}=?

Replacing:

75.5 kJ/mol=2* 90.25 kJ/mol + 0 - H_{NOCl}

Solving

-H_{NOCl}=75.5 kJ/mol - 2*90.25 kJ/mol

-H_{NOCl}=-105 kJ/mol

H_{NOCl}=105 kJ/mol

<u><em>The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol</em></u>

8 0
3 years ago
5. Write the names of the following ions.
Talja [164]
A.) Phosphate ion or Orthophosphate
d.) Hydroxide
D.) Ammonium
e.) Iron
C.) Nitrate
f.) Sulfur dioxide
5 0
2 years ago
In one experiment 7.62 g of Fe are allowed to react with 8.67 g of S
vichka [17]

calculate moles of both reagents given and the moles of FeS that each of them would form if they were in excess  

moles = mass / molar mass  

moles Fe = 7.62 g / 55.85 g/mol  

= 0.1364 moles  

1 mole Fe produces 1 mole FeS  

Therefore 7.62 g Fe can form 0.1364 moles FeS  

moles S = 8.67 g / 32.07 g/mol  

= 0.2703 moles S  

1 mole S can from 1 moles FeS  

So 8.67 g S can produce 0.2703 moles FeS  

The limiting reagent is the one that produces the least product. So Fe is limiting.  

The maximum amount of FeS possible is from complete reaction of all the limiting reagent.  

We have already determined that the Fe can form up to 0.1364 moles of FeS, so this is max amount of FeS you can get.  

Convert to mass

hope this helps :)

4 0
3 years ago
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