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Masteriza [31]
3 years ago
7

Which quantity or quantities must always be the same on both sides of a chemical equation? check all that apply. check all that

apply. the number of atoms of each kind the number of moles of each kind of molecule the sum of the masses of all substances involved the number of molecules of each kind?
Chemistry
2 answers:
Darya [45]3 years ago
6 0

Answer:

- The number of atoms of each kind.

- The sum of the masses of all substances involved.

Explanation:

Hello,

Chemical reactions are chemical processes wherein a series of reactants undergo a change in their composition due to a bond breaking caused by energetic means. In this manner, such change is constricted by two conditions which must be assured after the chemical reactions, they are:

- The number of atoms of each kind.

- The sum of the masses of all substances involved.

They are based on the law of conservation of mass which states that the mass can not be neither created nor destroyed, thus, the number of atoms of each kind and the sum of the masses must remain equal before and after the chemical reaction. For instance, consider the reaction:

2NaOH+H_2SO_4-->Na_2SO_4+2H_2O

Whereas the stoichiometric coefficients guarantees the first condition and the complete consumption-production the second one.

Best regards.

blagie [28]3 years ago
5 0

what are the answer choices?
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WEATHER PREDICTIONS FROM
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Answer:

These predictions are likely to be valid and should be needed.

Explanation:

Meteorologist made predictions about weather of a particular area on the data they received from the satellite. Meteorology is a branch of science in which we study about the atmosphere. It is mainly used for the forecasting of weather and meteorologists are those people or scientists who study the weather and research on different scales or instruments of weather.

7 0
3 years ago
A 5.00 gram sample of an oxide of lead PbxOy contains 4.33 g of lead. Determine simplest formula for the compund
Harrizon [31]

Answer: The empirical formula is PbO_2

Explanation:

Mass of Pb =  4.33 g

Mass of O = (5.00-4.33) g = 0.67 g

Step 1 : convert given masses into moles

Moles of Pb =\frac{\text{ given mass of Pb}}{\text{ molar mass of Pb}}= \frac{4.33g}{207g/mole}=0.021moles

Moles of O =\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{0.67g}{16g/mole}=0.042moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Pb = \frac{0.021}{0.021}=1

For O = \frac{0.042}{0.021}=2

The ratio of Pb O=  1: 2

Hence the empirical formula is PbO_2

6 0
3 years ago
For a particular isomer of C8H18, the combustion reaction produces 5113.3 kJ of heat per mole of C8H18(g) consumed, under standa
mart [117]

Answer: The standard enthalpy of formation of this isomer of C_8H_{18}(g) is -210.9 kJ

Explanation:

The given balanced chemical reaction is,

C_8H_{18}(g)+\frac{25}{2}O_2(g)\rightarrow 8CO_2(g)+9H_2O(g)

First we have to calculate the enthalpy of formation of  C_8H_{18}.

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{CO_2}\times \Delta H_f^0_{(CO_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{O_2}\times \Delta H_f^0_{(O_2)+n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})}]

where,

We are given:

\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H^o_f_{(O_2(g)))}=0kJ/mol

Putting values in above equation, we get:

-511.3kJ/mol=[(8\times -393.5)+(9\times -241.8)]-[(\frac{25}{2}\times 0)+(1\times \Delta H_f^0_{(C_8H_{18})}

\Delta H_f^0_{(C_8H_{18})}=-210.9kJ

7 0
4 years ago
Read 2 more answers
How many atoms are in 25.0 moles of calcium (Ca)?
arsen [322]

Answer:

1.8066e+24 is the answer

3 0
3 years ago
1. Perform calculations to determine the amount of 6.00x10-5 M stock solution needed to prepare 20.00 mL of 2.00x10-5 M dye solu
zhannawk [14.2K]

Answer:

1a. 6.70 ml of stock dye solution is required

1b. 10.0 ml of stock dye solution is required

1c. 4.00 ml of stock dye solution is required

Explanation:

1a. Using m₁v₁ = m₂v

6.00 * 10⁻⁵ * v₁ = 20.0 * 2.00 * 10⁻⁵

v₁ = 4 * 10⁻⁴/6.00 * 10⁻⁵

v₁ = 6.70 mL of stock solution

Therefore, 6.70 ml of stock dye solution is required

b. Using m₁v₁ = m₂v

2.00 * 10⁻⁵ * v₁ = 20.0 * 1.00 * 10⁻⁵

v₁ = 2 * 10⁻⁴/2.00 * 10⁻⁵

v₁ = 10.0 mL of stock solution

Therefore, 10.0 ml of stock dye solution is required

c. Using m₁v₁ = m₂v

1.00 * 10⁻⁵ * v₁ = 20.0 * 2.00 * 10⁻⁶

v₁ = 4 * 10⁻⁵/1.00 * 10⁻⁵

v₁ = 4.00 mL of stock solution

Therefore, 4.00 ml of stock dye solution is required

The procedure is then followed as in steps 2 to 4.

4 0
3 years ago
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