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3 years ago

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:

K+

because K is in IA (charge 1+)

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Which question would most likely be studied by a chemist?

lbvjy [14]

Answer:

Explanation:

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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:

O = 2 mol

H = 6 mol

C = 3 mol

Therefore the empirical formula of the compound is:

C3H6O2

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4 years ago

Baking soda and hydrochloric acid

olasank [31]

17 be cause i say so and if i dont help on this question i cant ask one

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3 years ago

Shtirlitz [24]

Answer:

I believe the answer to be period 3 group 1

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