All categories

3 years ago

PLS HELP MEEEEEEEEEEEE

Chemistry

1 answer:

Nikitich [7]3 years ago

6 0

Answer:

108- Hassium

11- sodium

56- Barium

92- uranium

25- manganese

103- lawrencium

10- neon

22- titanium

8- oxygen

14- silicon

32- germanium

39- yttrium

You might be interested in

What is H2O In chemistry

Serggg [28]

H2O, also known as water, stands for two hydrogen atoms and one oxygen atom. Hope this helps! ^^

5 0

3 years ago

Read 2 more answers

Which of these, all of which include earth, most likely contains the most dark matter

mixer [17]

Answer:

Plasma

Explanation:

8 0

3 years ago

each body cell of a goldfish contains 94 chromosomes.how many chromosomes are contained in a goldfish sex cell?

Shalnov [3]

47 I believe not 100% sure

8 0

3 years ago

The pressure of 5.00 L of gas increases from 1.50 atm to 1240 mmHg. What is the final volume of the gas, assuming constant tempe

Vsevolod [243]

(P1)(V1)=(P2)(V2)
(1.50)(5.00)=(1240/760)(V2)
(7.5)/(1240/760)=V2
V2=4.596774194 L

5 0

3 years ago

Read 2 more answers

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

PLS HELP MEEEEEEEEEEEE​

PLS HELP MEEEEEEEEEEEE