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WINSTONCH [101]
3 years ago
14

What is the charge on potassium (K) when it forms an ion? (1 point)

Chemistry
1 answer:
maks197457 [2]3 years ago
5 0

Explanation:

1+

K+

because K is in IA (charge 1+)

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Answer:

fd

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4 years ago
A sample of an unknown compound was decomposed and found to be composed of 1.36 mol oxygen, 4.10 mol hydrogen, and 2.05 mol carb
AfilCa [17]

We are given the number of moles:

O = 1.36 mol

H = 4.10 mol

C = 2.05 mol

 

To get the empirical formula, first divide everything by the smallest number of moles = 1.36 mol. So that:

O = 1 mol

H = 3 mol

C = 1.5 mol

 

Next step is to multiply everything by a number such that all will be a whole number. In this case, multiply by 2 to get a whole number for C, so that:

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8 0
4 years ago
Baking soda and hydrochloric acid
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4 0
3 years ago
1
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8 0
3 years ago
Just as the depletion of stratospheric ozone today threatens life on Earth today, its accumulation was one of the crucial proces
inessss [21]

Answer:

(a) rate = -(1/3) Δ[O₂]/Δt = +(1/2) Δ[O₃]/Δt  

(b) Δ[O₃]/Δt = 1.07x10⁻⁵ mol/Ls

Explanation:

By definition, t<u>he reaction rate for a chemical reaction can be expressed by the decrease in the concentration of reactants or the increase in the concentration of products:</u>    

aX → bY (1)

rate= -\frac{1}{a} \frac{\Delta[X]}{ \Delta t} = +\frac{1}{b} \frac{\Delta[Y]}{ \Delta t}

<em>where, a and b are the coefficients of de reactant X and product Y, respectively.        </em>

(a) Based on the definition above, we can express the rate of reaction (2) as follows:      

3O₂(g) → 2O₃(g) (2)    

rate = -\frac{1}{3} \frac{\Delta[O_{2}]}{\Delta t} = +\frac{1}{2} \frac{ \Delta[O_{3}]}{ \Delta t} (3)

(b) From the rate of disappearance of O₂ in equation (3), we can find the rate of appearance of O₃:  

rate = +\frac{1}{2} \frac{\Delta[O_{3}]}{ \Delta t} = -\frac{1}{3} \frac{\Delta[O_{2}]}{ \Delta t}

+\frac{\Delta[O_{3}]}{ \Delta t} = -\frac{2}{3} \frac{\Delta[-1.61 \cdot 10^{-5}]}{ \Delta t}          

\frac{\Delta[O_{3}]}{ \Delta t} = 1.07 \cdot 10^{-5} \frac{mol}{Ls}

So the rate of appearance of O₃ is 1.07x10⁻⁵ mol/Ls.

           

Have a nice day!

6 0
4 years ago
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