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Delvig [45]
2 years ago
11

PLEASE HELP ME!!

Chemistry
1 answer:
Mama L [17]2 years ago
5 0

Answer:

A

Explanation:

3 CARBON ATOMS 8 HYDROGEN

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How many grams of hydrogen gas get produced if 10 g of aluminum reacts?
Degger [83]

Answer:

9.8

Explanation:

8 0
2 years ago
Đốt cháy hết 9g kim loại magie Mg trong không khí tích tụ được 15g hợp chất magie oxit MgO. Biết rằng magie cháy là xảy ra phản
vfiekz [6]

Answer:

jijji[ojooooooooooooooooooooooooo

Explanation:

kjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj

5 0
2 years ago
What's the oxidation number of Li in LiNO3
Studentka2010 [4]

+1

Explanation:

To solve this problem, we need to set up an algebraic equation. Let us first understand the meaning of oxidation number.

The oxidation number is the formal charge assigned to an atom present in a molecule or formula unit

The algebraic sum of all oxidation numbers of atoms in an ion containing more than one kind of atom is the charge on the ion.

The algebraic sum of all oxidation number of atoms in a neutral compound is zero;

   The radical NO₃ has a formal charge of -1;

 let the oxidation number of Li = x

 x + (-1) = 0

  x = + 1

learn more:

Oxidation number brainly.com/question/10017129

#learnwithBrainly

4 0
2 years ago
From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp
Nina [5.8K]

<u>Answer:</u> The concentration of NH_3 required will be 0.285 M.

<u>Explanation:</u>

To calculate the molarity of NiC_2O_4, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of NiC_2O_4 = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M

For the given chemical equations:

NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}

Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9

Net equation: NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?

To calculate the equilibrium constant, K for above equation, we get:

K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48

The expression for equilibrium constant of above equation is:

K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}

As, NiC_2O_4 is a solid, so its activity is taken as 1 and so for C_2O_4^{2-}

We are given:

[[Ni(NH_3)_6]^{2+}]=0.016M

Putting values in above equations, we get:

0.48=\frac{0.016}{[NH_3]^6}}

[NH_3]=0.285M

Hence, the concentration of NH_3 required will be 0.285 M.

7 0
3 years ago
PLEASE HELP ME ASAP!!!!
VashaNatasha [74]

Answer:

16.   CsCl  =  Cs¹⁺ and Cl⁻¹

17.  CaF₂  =  Ca²⁺ and F⁻¹

18.   FeO  =  Fe²⁺ and O⁻²

19. Al₂O₃  =  Al³⁺ and O⁻²

4 0
2 years ago
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