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torisob [31]
2 years ago
8

I really need the help, come on please, I got a Walmart gift CARD for anyone who can fill this out correctly

Chemistry
1 answer:
Taya2010 [7]2 years ago
6 0

Answer:

1) 1 Na3PO4 + 3 KOH  3 NaOH + 1 K3PO4

2) 1 MgF2 + 1 Li2CO3  1 MgCO3 + 2 LiF

3) 1 P4 + 3 O2  2 P2O3

4) 2 RbNO3 + 1 BeF2  1 Be(NO3)2 + 2 RbF

5) 2 AgNO3 + 1 Cu  1 Cu(NO3)2 + 2 Ag

6) 1 CF4 + 2 Br2  1 CBr4 + 2 F2

7) 2 HCN + 1 CuSO4  1 H2SO4 + 1 Cu(CN)2

Explanation:

8) 1 GaF3 + 3 Cs  3 CsF + 1 Ga

9) 1 BaS + 1 PtF2  1 BaF2 + 1 PtS

10) 1 N2 + 3 H2  2 NH3

11) 2 NaF + 1 Br2  2 NaBr + 1 F2

12) 1 Pb(OH)2 + 2 HCl  2 H2O + 1 PbCl2

13) 2 AlBr3 + 3 K2SO4  6 KBr + 1 Al2(SO4)3

14) 1 CH4 + 2 O2  1 CO2 + 2 H2O

15) 2 Na3PO4 + 3 CaCl2  6 NaCl + 1 Ca3(PO4)2

16) 2 K + 1 Cl2  2 KCl

17) 2 Al + 6 HCl  3 H2 + 2 AlCl3

18) 1 N2 + 3 F2  2 NF3

19) 1 SO2 + 2 Li2Se  1 SSe2 + 2 Li2O

20) 2 NH3 + 1 H2SO4  1 (NH4)2SO4

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3 years ago
What molarity of nitric acid (HNO3) was used if 2.00 L must be used to prepare 4.5 L of a 0.25 M HNO3 solution?
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The   molarity  of (HNO₃) that was used  if 2.00 L must  be used   to prepare 4.5 L  of a 0.25M HNO₃ solution is   0.563 M


 <u><em>calculation</em></u>

  This is calculated  usind  M₁V₁=M₂V₂  formula

where,

         M₁(  molarity ₁) = ?

         V₁( volume ₁) = 2.00 L

        M₁ (molarity ₂) = 0.25M

         V₂( volume₂) = 4.5 L

make M₁ the subject  of the formula by  diving both side of the formula  by V₁

   M₁  is therefore = M₂V₂/V₁

M₁ =[ (0.25 M  x 4.5 L) / 2.00 L ]  =0.563 M

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