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Sliva [168]
3 years ago
14

According to Dalton, what happens to atoms during a chemical reaction? Atoms gain or lose energy. Atoms are rearranged. Old atom

s are destroyed. New atoms are created.
Chemistry
1 answer:
Hunter-Best [27]3 years ago
7 0

Answer:

Atoms are rearranged.

Explanation:

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Dogs can be acutely poisoned if they eat very small amounts of , a sweetener commonly found in sugar-free gum.
Advocard [28]

Answer:

Xylitol

Please give me the brainliest if you found this helpful

3 0
3 years ago
Which of the following factors can be changed for a gas without changing the mass of the gas? Select all that apply.
viva [34]
It is either
volume or pressure
8 0
3 years ago
Read 2 more answers
10.
Andru [333]

Answer: 1.8 moles Fe and 2.7 moles CO_2 are produced.

Explanation:

The balanced chemical reaction is:

Fe_2O_3+3CO\rightarrow 2Fe+3CO_2  

According to stoichiometry :

3 moles of CO require 1 mole of Fe_2O_3

Thus 2.7 moles of CO will require=\frac{1}{3}\times 2.7=0.9moles  of Fe_2O_3

Thus CO is the limiting reagent as it limits the formation of product and Fe_2O_3 is the excess reagent.

As 3 moles of CO give = 2 moles of Fe  and 3 moles of CO_2

Thus 2.7 moles of CO will give =\frac{2}{3}\times 2.7=1.8moles  of Fe  and \frac{3}{3}\times 2.7=2.7moles  of CO_2

Thus 1.8 moles Fe and 2.7 moles CO_2 are produced.

7 0
3 years ago
Science.... <br> homework!
Leokris [45]
The answer to this question is 2 and 3
5 0
3 years ago
The value of ka for nitrous acid (hno2) at 25 ∘c is 4.5×10−4. what is the value of δg at 25 ∘c when [h+] = 5.9×10−2m , [no2-] =
LuckyWell [14K]
To get the value of ΔG we need to get first the value of ΔG°:

when ΔG° = - R*T*㏑K

when R is constant in KJ = 0.00831 KJ

T is the temperature in Kelvin = 25+273 = 298 K

and K is the equilibrium constant = 4.5 x 10^-4

so by substitution:

∴ ΔG° = - 0.00831 * 298 K * ㏑4.5 x 10^-4

        = -19 KJ

then, we can now get the value of ΔG when:

ΔG = ΔG° - RT*㏑[HNO2]/[H+][NO2]

when ΔG° = -19 KJ

and R is constant in KJ = 0.00831 

and T is the temperature in Kelvin = 298 K

and [HNO2] = 0.21 m & [H+] = 5.9 x 10^-2 & [NO2-] = 6.3 x 10^-4 m  

so, by substitution:

ΔG = -19 KJ - 0.00831 * 298K* ㏑(0.21/5.9x10^-2*6.3 x10^-4 )

       = -40

     
8 0
3 years ago
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