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3 years ago

Mendeleev arranged the elements by what physical property? atomic number atomic mass boiling point density

Chemistry

2 answers:

____ [38]3 years ago

5 0

Answer : The correct option is, Atomic mass.

Explanation :

According to Mendeleev, the elements and their compounds having same chemical properties showed a periodic trend. He arranged the elements having similar properties below each other into groups.

He arranged the elements in order of their increasing relative atomic masses.

Hence, the correct answer is, Atomic mass.

m_a_m_a [10]3 years ago

3 0

Mendeleev arranged the elements according to their atomic masses

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2 years ago

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Describe un modelo matemático para representar lo que sucede en cada una de las reacciones químicas (incluye toda la simbología

GuDViN [60]

Responder:

2H2 + O2 → 2H2O

CaO + H2O → Ca (OH) 2

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Explicación:

2H2 + O2 → 2H2O

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CaO + H2O → Ca (OH) 2

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2 years ago

Which evidence is an indication that a chemical reaction has occurred?

Luden [163]

Answer:

Some signs of a chemical change are a change in color and the formation of bubbles. The five conditions of chemical change: color change, formation of a precipitate, formation of a gas, odor change, temperature change.

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2 years ago

Hydrazine (N2H4) is a fuel used by some spacecraft. It is normally oxidized by N2O4 according to the following equation: N2H4(l)

vitfil [10]

Answer:

The enthalpy of the reaction is coming out to be -380.16 kJ.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

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:

$N_2H_4(l)+N_2O_4(g)\rightarrow 2N_2O(g)+2H_2O(g)$

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

$\Delta H_{rxn}=[(2 mol\times \Delta H_f_{(N_2O)})+(2 mol\times\Delta H_f_{(H_2O)} )]-[(1 mol\times \Delta H_f_{(N_2H_4)})+(1 mol\times \Delta H_f_{(N_2O_4)})]$

We are given:

$\Delta H_f_{(N_2O)}=81.6 kJ/mol\\\Delta H_f_{(H_2O)}=-241.8 kJ/mol\\\Delta H_f_{(N_2H_4)}= 50.6 kJ/mol\\\Delta H_f_{(N_2O_4)}=9.16 kJ/mo$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(2 mol\times 81.6 kJ/mol)+2 mol\times -241.8 kJ/mol)]-[(1 mol\times (50.6 kJ/mol))+(1 mol\times (9.16))]\\\\\Delta H_{rxn}=-380.16 kJ$

Hence, the enthalpy of the reaction is coming out to be -380.16 kJ.

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3 years ago

Help pleas I dont understand

Temka [501]

The *independent* variable is the one that you always can control. However, the dependent variables are not since they **depend** on another.

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