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3 years ago

2 drops of vinegar & 2 drops ammonia solution or not

Chemistry

1 answer:

Naya [18.7K]3 years ago

5 0

Answer:

No, this is false because we know about the law that says if you add two things together you are adding the masses together, 2 drops of vinegar and 2 drops of ammonia would equal four drops of solution not two

Hope this helps!!

Explanation:

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In the winter, a heated home in the Northeast might be maintained at a temperature of 78°F. What is this temperature on the Cels

masya89 [10]

Answer: $25.6^0C$ , $298.6K$

Explanation:

Temperature of the gas is defined as the degree of hotness or coldness of a body. It is expressed in units like $^0C$ and $K$

These units of temperature are inter convertible.

We are given:

Temperature of the gas = $78^0F$

Converting this unit of temperature into $^0C$ by using conversion factor:

$^oC=\frac{5}{9}\times (^oF-32)$

$^0C=\frac{5}{9}\times (78^oF-32)$

$25.6^0C$

Converting this unit of temperature into $K$ by using conversion factor:

$K=t^0C+273$

$K=25.6+273$

$298.6K$

Thus the temperature on the Celsius and Kelvin scales are $25.6^0C$ and $298.6K$ respectively.

6 0

3 years ago

An excited ozone molecule, O3*, in the atmosphere can undergo one of the following reactions,O3* → O3 (1) fluorescenceO3* → O +

Maurinko [17]

Answer:

The simplified expression for the fraction is $\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$

Explanation:

From the given information:

O3* → O3 (1) fluorescence

O + O2 (2) decomposition

O3* + M → O3 + M (3) deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

where cM is the concentration of the inert molecule

The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

$\text {X} = \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) } } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{ {(k_1 \times cO) +(k_2 \times cO) + (k_3 \times cO \times cM) } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3 \times cM) }$

$\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$ since cM is the concentration of the inert molecule

7 0

3 years ago

Give the symbol for the element with the following orbital diagram:

wel

Answer: 0%

Explanation:

You got a bad grade

5 0

3 years ago

What information is need to calculate the percent composition of a compound?

Anestetic [448]

Answer:

Molecular formula

Explanation:

Molecular formula in the first place is required to understand which compound we have. We then should refer to the periodic table and find the molecular weight for each atom. Adding individual molecular weights together would yield the molar mass of a compound.

Then, dividing the total molar mass of a specific atom by the molar mass of a compound and converting into percentage will provide us with the percentage of that specific atom.

E. g., calculate the percent composition of water:

molecular formula is $H_2O$ ;
calculate its molar mass: [tex]M = 2M_H + M_O = 2\cdot 1.00784 g/mol + 16.00 g/mol = 18.016 g/mol;
find the percentage of hydrogen: [tex]\omega_H = \frac{2\cdot 1.00784 g/mol}{18.016 g/mol}\cdot 100 \% = 11.19 %;
find the percentage of oxygen: [tex]\omega_O = \frac{16.00 g/mol}{18.016 g/mol}\cdot 100 \% = 88.81 %.

8 0

3 years ago

2. This graph shows the energy involved in a reaction involving two molecules. Describe what is happening or what the state of t

blagie [28]

If you think of it endothermic is when there is energy needed for the reaction to occur and exothermic is when the reaction releases energy

6 0

3 years ago

2 drops of vinegar & 2 drops ammonia solution or not​

2 drops of vinegar & 2 drops ammonia solution or not