Hi . I need help on this problem please help.

Which particle is most likely present in the nucleus of the atom

k0ka [10]

Oh protons and neutrons

8 0

3 years ago

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Scott and James work at a grocery store. After the grocery

sashaice [31]

Answer: b) Crash 2; the force on the cart was stronger in this crash, so the force on the skateboard was also stronger.

Explanation:

7 0

3 years ago

Which of these describes an exothermic reaction?

Paul [167]

A nswer: -

C. Energy is released by the reaction

Explanation:-

An exothermic reaction is one in which during the progress of the reaction heat is evolved.

So energy is released by the reaction.

It cannot be created as energy is neither created nor destroyed as per the Law of conservation of energy. Energy is not transferred either.

The energy released during the progress of the reaction originates from the chemical bonds of the reactants as they break during their conversion into products.

3 0

2 years ago

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What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )

yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution : Given,

Concentration (c) = 0.150 M

Acid dissociation constant = $k_a=4.5\times 10^{-4}$

The equilibrium reaction for dissociation of $HNO_2$ (weak acid) is,

$HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+$

initially conc. c 0 0

At eqm. $c(1-\alpha)$ $c\alpha$ $c\alpha$

First we have to calculate the concentration of value of dissociation constant $(\alpha )$ .

Formula used :

$k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

Now put all the given values in this formula ,we get the value of dissociation constant $(\alpha )$ .

$4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}$

$4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2$

$0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0$

By solving the terms, we get

$\alpha=0.0533$

No we have to calculate the concentration of hydronium ion or hydrogen ion.

$[H^+]=c\alpha=0.150\times 0.0533=0.007995 M$

Now we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (0.007995 M)$

$pH=2.097\approx 2.1$

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0

3 years ago

Calculate moles of oxygen atoms in 0.68mol of KMnO4

Dominik [7]

Alright, so that means we have 0.68 mol of the compound

For each 1 mol of the compound, we have 4*1 oxygens (because there are four oxygens in the formula)
Therefore for each 0.68 mol of the compound, we have 4*0.68 moles of oxygen!

6 0

3 years ago