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3 years ago

1. What determines the reactivity of an atom?

1 answer:

5 0

The number of electrons in the outermost shell of an atom determines its reactivity.

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How does air bubbles affect density of a solid object?

kirill [66]

it decreases the density of the object the air bubbles take up space. it increases the volume of the object slightly but the objects weight remains the same, hence the objects density decreases

6 0

3 years ago

How many liters of oxygen gas are there in 45.0g O, at 0 degrees C and 1 atm of pressure? make sure to round to the correct numb

ZanzabumX [31]

Answer:

You can use the pdf to help you

Explanation:

Download pdf

3 0

2 years ago

From goal line to goal line, a football field is 300 ft long. If a player catches the ball while standing on one goal line, runs

KATRIN_1 [288]

Answer:

The player ran 91.44m.

Explanation:

The problem gives you the total distance between goal line to goal line in feet, and the answer must be given in meters, so you should convert the distance the player run from ft to m, because the player run the same distance from goal line to goal line to scores the touchdown.

So, you should apply the following conversion factor:

$300ft*\frac{0.3048m}{1ft}= 91.44m$

The player ran 91.44m.

3 0

3 years ago

The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be

Mrac [35]

Answer: The freezing point of solution is 5.35°C

Explanation:

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 4.90°C/m

$m_{solute}$ = Given mass of solute (naphthalene) = 2.60 g

$M_{solute}$ = Molar mass of solute (naphthalene) = 128.2 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC$

Hence, the freezing point of solution is 5.35°C

3 0

3 years ago

Consider the diagram below

Artyom0805 [142]

Energy of the reactants would be correct

5 0

2 years ago