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Zolol [24]
3 years ago
13

What do all the atomic models have in common?

Chemistry
1 answer:
Viefleur [7K]3 years ago
4 0
All atomic models have a nucleus
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During fusion
Veronika [31]

Answer:

A

Explanation:

uranium atoms are fused together

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The independent variable is the only factor that a scientist
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Changes. the dependent variable is affected by the independent variable
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If there are 10 navy beans,3 pinto beans,and 17 lentils in a container,what is the percent composition of the container by bean?
Stells [14]

The percent abundance in the container by bean will be 43.33 %  if there are 10 navy beans, 3 pinto beans, and 17 lentils in a container

<h3>What is Percentage composition ?</h3>

The percentage composition of a given compound is defined as the ratio of the amount of each element to the total amount of individual elements present in the compound multiplied by 100.

% composition by bean = number of beans / total number of objects in container x 100

  • Total object = 10 ( navy beans) + 3 (pinto beans) + 17 ( lentils) = 30 objects
  • Total beans = 10 ( navy beans) + 3 (pinto beans) = 13 beans

hence ;

  • % composition by beans = 13/30 x 100 = 43.33 %

Therefore, the percent composition of the container by bean is 43.33 %

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7 0
2 years ago
I really need help ! This is science
lina2011 [118]

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7. genetics because hereditary is not the study of hereditary.

8. is hereditary because it is the study of how things pass on to next generation what is inherited.

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3 years ago
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Calculate ΔHo for the following reaction ussing the given bond dissociation energiesCH4(g) + 2O2(g) --&gt; CO2(g) + 2H2O(g)BOND
Mazyrski [523]

Answer:

The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,

Explanation:

CH_4(g) + 2O_2(g)\rightarrow CO_2(g) + 2H_2O(g) ,ΔH° = ?

We are given with:

\Delta H_{O-O}=142 kJ/mol

\Delta H_{O=O}=498 kJ/mol

\Delta H_{H-O}=459 kJ/mol

\Delta H_{C-H}=411 kJ/mol

\Delta H_{C-O}=358 kJ/mol

\Delta H_{C=O}=799 kJ/mol

ΔH° =  

(Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)

\Delta H^o=(1 mol\times 4\times \Delta H_{C-H}+2 mol\times 1\times \Delta H_{O=O})-(1 mol\times 2\times \Delta H_{C=O}+2 mol\times 2\times\Delta H_{H-O})

\Delta H^o=(1 mol\times 4\times 411 kJ/mol+2 mol\times 1\times 498 kJ/mol)-(1 mol\times 2\times 799 kJ/mol+2 mol\times 2\times 459 kJ/mol)

\Delta H^o=-794kJ

\Delta H^o>0 endothermic reaction

\Delta H^o exothermic reaction

The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,

4 0
4 years ago
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