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Semenov [28]
3 years ago
14

Protons are positively charged and repel other protons. Which other particle is found in the nucleus and separates protons so th

at the strong force can hold the nucleus together?
Chemistry
1 answer:
VashaNatasha [74]3 years ago
8 0
Neutrons which doesn't have any charge
You might be interested in
Help me plzz in chemistry​
AlexFokin [52]

a) f2 loses electrons and is oxidised. li gains electrons and is reduced

c)sn2+ loses electrons and is oxidised.Al gains electrons and is reduced

b)br2 loses electrons and is oxidised. I- in KI gains electrons and is reduced

5 0
3 years ago
At 400 K oxalic acid decomposes according to the reaction:H2C2O4(g)→CO2(g)+HCOOH(g)In three separate experiments, the intial pre
padilas [110]

Answer:

v = 2,66x10⁻⁵ P[H₂C₂O₄]

Explanation:

For the reaction:

H₂C₂O₄(g) → CO₂(g) + HCOOH(g)

At t = 0, the initial pressure is just of H₂C₂O₄(g). At t= 20000 s, pressures will be:

H₂C₂O₄(g) = P₀ - x

CO₂(g) = x

HCOOH(g) = x

P at t=20000 is:

P₀ - x + x + x = P₀+x. That means P at t=20000s - P₀ = x

For 1st point:

x = 92,8-65,8 = 27

Pressure of H₂C₂O₄(g) at t=20000s: 65,8-27 = 38,8

2nd point:

x = 130-92,1 = 37,9

H₂C₂O₄(g): 92,1 - 37,9 = 54,2

3rd point:

x = 157-111 = 46

H₂C₂O₄(g): 111-46 = 65

Now, as the rate law is :

v = k P[H₂C₂O₄]

Based on integrated rate law, k is:

(- ln P[H₂C₂O₄] + ln P[H₂C₂O₄]₀) / t = k

1st point:

k = 2,64x10⁻⁵

2nd point:

k = 2,65x10⁻⁵

3rd point:

k = 2,68x10⁻⁵

The averrage of this values is:

k = 2,66x10⁻⁵

That means law is:

v = 2,66x10⁻⁵ P[H₂C₂O₄]

I hope it helps!

4 0
3 years ago
Calculate the molar concentration of the Br⁻ ions in 0.51 M MgBr2(aq), assuming that the dissolved substance dissociates complet
Y_Kistochka [10]
MgBr2(aq) is an ionic compound which will have the releasing of 2 Br⁻ ions ions in water for every molecule of MgBr2 that dissolves.
MgBr2(s) --> Mg+(aq) + 2 Br⁻(aq)
            [Br⁻] = 0.51 mol MgBr2/1L × 2 mol Br⁻ / 1 mol MgBr2 = 1.0 M
The answer to this question is [Br⁻] = 1.0 M
6 0
3 years ago
Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp
gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200  

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in  j/mol R Ea is

Ea=35.5*1000 j/mol R

ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

Read more about rate constant

brainly.com/question/20305871

#SPJ1

5 0
2 years ago
Someone pls help me I will make you brain
wlad13 [49]

Answer:

It is true answer. Ozone = O3

7 0
3 years ago
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