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3 years ago

Imagine that instead of balls, you were rolling positively charged ions of the same element).

1 answer:

8 0

The ion with a +3 charge would be deflected the most by the magnet because it has the strongest positive charge therefore it will be the one being the most repelled. and the ion that would be deflected the least would be the ion with a +1 charge because it has the least amount of charge so the magnet will still repel it but just not as much as it will repel the ion with a +3 charge.

hope that helps

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A gold nugget has a density of 38.6g/cm3 and a mass of 270.2. what is its volume?

aivan3 [116]

Density is given by the equation D=m/V, were D is density, m is mass in grams, and V is volume in cubic centimeters.

In this problem, we have density and we have mass so we can plug into the equation and solve for V.

38.6=270.2/V

*Multiply both sides by V*

38.6V=270.2

*Divide both sides by 38.6*

V=7

The volume of the gold nugget is 7cm3.

Hope this helps!!

5 0

3 years ago

Read 2 more answers

A container of argon with a volume of 22 ml has a pressure of 101.0 kPa.

Alexeev081 [22]

Answer:

The new pressure is 44.4 kPa.

Explanation:

We have,

Initial volume, $V_1=22\ ml$

Initial pressure, $P_1= 101.0\ kPa$

It is required to find the new pressure when the volume is increased to 50 ml. The relationship between pressure and volume is known as Boyle's law.

$PV=k\\\\P_1V_1=P_2V_2$

$P_2$ is final pressure

$P_2=\dfrac{P_1V_1}{V_2}\\\\P_2=\dfrac{22\times 101\times 10^3}{50}\\\\P_2=44440\ Pa\\\\P_2=44.4\ kPa$

So, new pressure is 44.4 kPa.

6 0

3 years ago

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.32 g of ethane is

Pie

Answer:

There will remain 8.06 grams of ethane

Explanation:

Step 1: Data given

Mass of ethane = 9.32 grams

Mass of oxygen = 12.0 grams

Molar mass ethane = 30.07 g/mol

Molar mass oxygen = 32.00 g/mol

Step 2: The balanced equation

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

Step 3: Calculate moles ethane

Moles ethane = mass ethane / molar mass ethane

Moles ethane = 9.32 grams / 30.07 g/mol

Moles ethane = 0.3099 moles

Step 4: Calculate moles oxygen

Moles oxygen = 12.0 grams / 32.0 g/mol

Moles oxygen = 0.375 moles

Step 5: Calculate the limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.375 moles)

Ethane is in excess. There will react 2/7 * 0.375 = 0.107 moles

There will remain 0.375 - 0.107 = 0.268 moles

Step 6: Calculate mass ethane

Mass ethane = moles ethane * molar mass ethane

Mass ethane = 0.268 moles * 30.07 g/mol

Mass ethane = 8.06 grams

There will remain 8.06 grams of ethane

7 0

3 years ago

Which is the term for how vegetation influences precipitation?

Scilla [17]

The answer to this question is A- evaporation

3 0

3 years ago

Molar mass of Ag2SO4 Full work Pls

IceJOKER [234]

This would be the molar mass.

5 0

3 years ago