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salantis [7]
2 years ago
14

Iron has a density of 7.87 g/cm3. What is the volume in cm3 of 3.729 g of iron?

Chemistry
1 answer:
KIM [24]2 years ago
6 0

If iron has a density of 7.87g/cm³ and a mass of 3.729g, then the volume of iron is 0.474cm³

HOW TO CALCULATE VOLUME:

  • The volume of a substance can be calculated by dividing the mass by its density. That is;

Volume (mL) = mass (g) ÷ density (g/mL)

  • The density of iron is given as 7.87g/cm³ while its mass is 3.729g of iron. Hence, the volume can be calculated as follows:

Volume = 3.729 ÷ 7.87

Volume = 0.474cm³

Therefore, the volume of iron is 0.474cm³

Learn more: brainly.com/question/2040396?referrer=searchResults

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Signs of a chemical change
dezoksy [38]

Answer: There are five signs of chemical change

  • Color Change
  • Production of an odor
  • Change of Temperature
  • Evolution of Gas (bubbles start to form)
  • Precipitate (starts to form a solid)

When these signs start to form you know chemical change is at work.

Hope this helps :)

5 0
3 years ago
Read 2 more answers
The individual dipole moments in ammonia (NH3) do not cancel each other
kozerog [31]

Answer:

                    The strongest force that exists between molecules of Ammonia is <em>Hydrogen Bonding</em>.

Explanation:

                    Hydrogen Bond Interactions are those interactions which are formed between a partial positive hydrogen atom bonded directly to most electronegative atoms (i.e. F, O and N) of one molecule interacts with the partial negative most electronegative atom of another molecule.

                    Hence, in ammonia the nitrogen atom being more electronegative element than Hydrogen will be having partial negative charge and making the hydrogen atom partial positive. Therefore, the attraction between these partials charges will be the main force of interaction between ammonia molecules.

                  Other than Hydrogen bonding interactions ammonia will also experience dipole-dipole attraction and London dispersion forces.

4 0
3 years ago
Gamma rays are often used to kill microorganisms in food, in an attempt to make the food safer. Some people contend that this ir
kumpel [21]
<span>While treating food with gamma rays kills microorganism by damaging their DNA, the energy of gamma rays rips off electrons from atoms hence ionizing them (causing free radicals). However, gamma rays do not make the food atoms radioactive. The body has a natural mechanism of riding the body of free radicals. However, large quantities of radicals in the body can cause damage.</span>




7 0
3 years ago
When you apply 1000 joules of energy to 50 grams of water its temperature changes to 30 degrees . What was the initial temperatu
expeople1 [14]

Answer:

25.2°C

Explanation:

Given parameters:

Energy applied to the water  = 1000J

Mass of water  = 50g

Final temperature  = 30°C

Unknown:

Initial temperature  = ?

Solution:

To solve this problem, we use the expression below:

            H  = m c Ф

H is the energy absorbed

m is the mass

c is the specific heat capacity

Ф is the change in temperature

     1000  = 50 x 4.184 x (30  -  initial temperature )

     1000  = 209.2(30 - initial temperature)

      4.78  = 30 - initial temperature

      4.78  - 30  = - initial temperature

         Initial temperature  = 25.2°C

7 0
3 years ago
describe in general terms an experiment to determine the molal freezing point depression constant kf of water. Assume the availa
Dvinal [7]
A solution (in this experiment solution of NaNO₃) freezes at a lower temperature than does the pure solvent (deionized water). The higher the solute concentration (sodium nitrate), freezing point depression of the solution will be greater.
Equation describing the change in freezing point: 
ΔT = Kf · b · i.
ΔT - temperature change from pure solvent to solution.
Kf - the molal freezing point depression constant.
b -  molality (moles of solute per kilogram of solvent).
i - Van’t Hoff Factor.
First measure freezing point of pure solvent (deionized water). Than make solutions of NaNO₃ with different molality and measure separately their freezing points. Use equation to calculate Kf.

6 0
3 years ago
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