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german
3 years ago
12

How to show work ???(NH4)3PO4 + Pb(NO3)4 -> Pb3(PO4)4 + NH4NO3

Chemistry
1 answer:
Novay_Z [31]3 years ago
6 0

Explanation: We are given a chemical reaction and we need to balance it.

Balancing a chemical reaction means that the total mass on the reactant side will be equal to the total mass on the product side. For this, we need to balance the number of atoms on both the sides.

The given balanced chemical reaction follows:

4(NH_4)_3PO_4+3Pb(NO_3)_4\rightarrow Pb_3(PO_4)_4+12NH_4NO_3

By Stoichiometry,

4 moles of Ammonium phosphate reacts with 3 moles of lead (IV) nitrate to produce 1 mole of lead (IV) phosphate and 12 moles of ammonium nitrate.

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Balance the following equation;C<img src="https://tex.z-dn.net/?f=C5H12%28g%29%2BO2%28g%29%3DC02%28g%29%2BH2O%28l%29" id="TexFor
Bumek [7]

put 8 in front of the oxygen in the reactants side to make it 16 molecules then put a 5 in front of the co2 in the product side to balance the carbon atoms then put a 6 in front of the H20 on the product side this balances both the hydrogen and oxygen atoms here is a representation

C5H12(g)+8O2(g)=5CO2(g)+6H20

4 0
2 years ago
How is a crystal different from a solid made of individual molecules
Dmitry [639]
In a crystal, the molecules are closer together as they are in any solid. they have less room to move, and might even be combined together rather than individual

3 0
2 years ago
I didn’t mean to click that answer I dropped my phone on the computer.
Yanka [14]

its A plant like protist produce oxygen ...

5 0
3 years ago
Correct answer please i need it before 12:07
choli [55]

Answer:

xenon

Explanation:

Xenon is a non-metal, odorless gas, consist of single atoms, colorless, and 8 valence electrons out of everything else on this list.

6 0
2 years ago
At equilibrium, the concentrations of the products and reactants for the reaction, H2 (g) + I2 (g)  2 HI (g), are [H2] = 0.106
lana [24]

Answer:

The new equilibrium concentration of HI: <u>[HI] = 3.589 M</u>          

Explanation:

Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M        

Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M

<em>Given chemical reaction:</em> H₂(g) + I₂(g) → 2 HI(g)

The equilibrium constant (K_{c}) for the given chemical reaction, is given by the equation:

K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}

<u><em>At the original equilibrium state:</em></u>

K_{c} = \frac {(1.29\: M)^{2}}{(0.106\: M) \times (0.022\: M)}

K_{c} = \frac {1.6641}{0.002332} = 713.59

<u><em>Therefore, at the new equilibrium state:</em></u>

K_{c} = \frac {[HI]^{2}}{(0.95\: M) \times (0.019\: M)}

\Rightarrow K_{c} = 713.59 = \frac {[HI]^{2}}{0.01805}

\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88

\Rightarrow [HI] = \sqrt {12.88} = 3.589 M

<u>Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M</u>

6 0
2 years ago
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