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2 years ago

14

Which two subatomic particles are located in the nucleus of an atom?

1 answer:

natta225 [31]2 years ago

4 0

Two subatomic particles that are located in the nucleus of an atom are Protons and Neutrons.

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In a particular genetic cross, the probability of the offspring having a certain trait is Which percentage represents this proba

AnnyKZ [126]

I think it is 75% sorry if I am wrong but if I am correct can I get brainlest?

8 0

3 years ago

For the reaction2 C4H10 (g) + 13 O2 (g) → 8 CO2 (g) + 10 H2O (g)ΔH° is -125 kJ/mol and ΔS° is +253 J/K ∙ mol. This reaction is _

ivanzaharov [21]

Answer:

C. spontaneous at all temperatures

Explanation:

The spontaneity of reaction is determined by the sign of the gibbs free energy.

A negative sign denotes that the reaction is spontaneous, positive sign means the reaction is not spontaneous.

From the question;

ΔS° = +253 J/K

ΔH° = -125 kJ/mol

ΔG = ΔH° - TΔS°

From the data given, the condition in which we can obtain a negative value of G, is at any value of T.

For any value of T, G would always be a negative value.

This means the correct option is option C.

3 0

3 years ago

A scientist who is trying to develop a material that is soft and malleable and also conducts electricity will likely use a__

Sonbull [250]

It's A, metals. Metals are solid at room temperature but still very malleable and the best conductors. Metalloids can still conduct electricity but nowhere near as well as metals.

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3 years ago

A glass jar is sealed, trapping air inside. The jar warms up after it is left ourside in the sun. What will happpen to the press

amid [387]

Answer:

the particles start to spread out and cause it to get warmer

Explanation:

7 0

3 years ago

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

5 0

3 years ago