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tino4ka555 [31]
2 years ago
13

Identify how energy changed form so it could be use in each example.

Chemistry
1 answer:
Klio2033 [76]2 years ago
6 0

Answer:

→A space heater warms a room⇒By converting electrical energy into heat energy.

→You eat dinner⇒Our  body uses chemical energy to break down the food

→A radio plays⇒By converting electrical energy into sound energy in an electric radio. And by converting chemical energy into sound energy in a radio which runs with a cell.

Explanation:

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When 28 g of nitrogen and 6 g of hydrogen react, 34 g of ammonia are produced. if 100 g of nitrogen react with 6 g of hydrogen,
vampirchik [111]
N2 + 3H2 --> 2NH3
When 100g of N2 , no of moles of N2= 100/(28)=3.57 mol
no. of moles of h2 = 6/(2)=3mol
Therefore h2 is limiting reagent.
no. of moles of ammonia= 3/3*2=2moles
mass of ammonia produced= 2 mol * (14+3)= 34g
4 0
3 years ago
Read 2 more answers
What is an example of kinetic energy at work,1 a ball lodge in the tree, 1 a frisbee flaying in the air, 3 a car in the garage
AfilCa [17]

Answer:

a frisbee flaying in the air

Explanation:

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

K.E = \frac{1}{2}MV^{2}

Where;

  • K.E represents kinetic energy measured in Joules.
  • M represents mass measured in kilograms.
  • V represents velocity measured in metres per seconds square.

Hence, an example of kinetic energy at work is a frisbee flaying in the air because it would possess energy due to its motion in the air.

8 0
3 years ago
A compound is found to contain 43.66% of P and 56.33% of O.
jek_recluse [69]

Answer: P2O5 is the empirical formula.

Explanation: When given percentages you can assume that many grams of each atom are in the compound. Then you divide grams by the molar mass of each element, giving you moles. Once you have moles, divide by the smaller molar amount, which should give you 1 mol of Phosphorus and 2.5 mol of Oxygen. Then multiply by 2 in order for both moles to be a whole number. This gets you 2 and 5.

6 0
3 years ago
The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +
Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b) The solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

Explanation:

(a)  Chemical equation for the given reaction in pure water is as follows.

           CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)

Initial:                         0            0

Change:                    +x           +x

Equilibm:                   x             x

K_{sp} = 1.2 \times 10^{-6}

And, equilibrium expression is as follows.

          K_{sp} = [Cu^{+}][Cl^{-}]

       1.2 \times 10^{-6} = x \times x

             x = 1.1 \times 10^{-3} M

Hence, the solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b)  When NaCl is 0.1 M,

       CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq),  K_{sp} = 1.2 \times 10^{-6}

   Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq),  K = 8.7 \times 10^{4}

Net equation: CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

               K' = K_{sp} \times K

                          = 0.1044

So for, CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

Initial:                     0.1                 0

Change:                -x                   +x

Equilibm:            0.1 - x                x

Now, the equilibrium expression is as follows.

              K' = \frac{CuCl_{2}}{Cl^{-}}

         0.1044 = \frac{x}{0.1 - x}

              x = 9.5 \times 10^{-3} M

Therefore, the solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

7 0
3 years ago
Which elements are metalloids ? Check all that apply
Degger [83]

Answer:

here is definition with example

8 0
3 years ago
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