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stira [4]
3 years ago
11

An isotope will lose nucleons,protons,or neutrons in order to become more

Chemistry
2 answers:
Debora [2.8K]3 years ago
8 0
Many atoms if they are radioactive isotopes will loss protons and neutrons as radiation in order to gain and become more stable 

hope that helps
RoseWind [281]3 years ago
3 0

Answer:

to become MORE STABLE :)

Explanation:

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Ce cantitate de O se gaseste in 5,8 g hidroxid de magneziu
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Answer:

separare pâlnii de picurare balon cotat. 1. 2. 3. 4 5 6 7. 8. 9. 1 ... Mod de lucru: 25 g Na2SO4∙7H2O se dizolvă în cantitatea minim ... Exemple: NaOH – hidroxid de sodiu.

Explanation:

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4 years ago
What describes the solvent in any solution
Lady_Fox [76]
Solvent is more than a solute. Like salt water. Water is the solvent and salt will be the solute
5 0
3 years ago
Read 2 more answers
The ___ blends into outer space.
Paraphin [41]

Answer:

Exosphere

Explanation:

it is found at the end reaching outer space

3 0
3 years ago
2C4 H10 + 13O2 -→ 8CO2 + 10H2O If 356 moles of O2 are available to react, how many moles of CO2 will fom?
nlexa [21]

Answer:

Answer:

The mole ratio of C₄H₁₀ and CO₂ is 2 : 8, which  simplifies to 1 : 4.

Explanation:

The mole ratio is the relative proportion of the moles of products or reactants that participate in the reaction according to the chemical equation.

The chemical equation given is:

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

Once you check that the equation is balanced, you can set the mole ratios for all the reactants and products. The coefficients used in front of each reactant and product, in the balanced chemical equation, tells the mole ratios.

In this case, they are: 2 mol C₄H₁₀ :  13 mol O₂ : 8 mol CO₂ : 10 mol H₂O

Since you are asked about the mole ratio of C₄H₁₀ and CO₂ it is:

2 mol C₄H₁₀ : 8 mol CO₂ , which dividing by 2, simplifies to

1 mol C₄H₁₀ : 4 mol CO₂, or

1 : 2.

Explanation:

5 0
3 years ago
I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol
galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = \Delta H_f_{(I_2)}=62.438 kJ/mol

Enthalpy of formation of chlorine gas = \Delta H_f_{(Cl_2)}=0 kJ/mol

Enthalpy of formation of ICl gas = \Delta H_f_{(ICl)}=17.78 kJ/mol

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]

=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol

Enthaply change when 1.62 moles of iodine gas recast:

\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ

Entropy of the surrounding = \Delta S^o_{surr}=\frac{\Delta H}{T}

=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

4 0
3 years ago
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