Why doesn't the density of the object change when you adjust the mass and volume of the objects?

1 answer:

6 0

Because as you change the mass the volume will change and as you change the volume the mass will change too, it's a constant ratio for each element. Mass and volume are not factors that affect the density. factors that affect the density are temperature and type of the material/element

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Why are cars manufactures exploring hydrogen fuel cell technology as a possiable replacement for gasoline burning engines

horsena [70]

To possibly take down on the pollution we produce in the world

4 0

3 years ago

How much heat in joules is gained when a 34.5 g bar of gold is heated from 26.4°C to 75.3°C ?specific heat of gold on 0.129 J/gr

Sati [7]

Answer:

Q = 217.63J

Explanation:

Data;

Mass = 34.5g

T1 = 26.4°C

T2 = 75.3°C

c = 0.129J/g°C

Energy (Q) = mc(T2 - T1)

Q = 34.5 * 0.129 *(75.3 - 26.4)

Q = 4.4505 * 48.9

Q = 217.629J

Q = 217.63J

The heat gained was 217.63J

7 0

3 years ago

In the reaction below, what is the limiting reactant when 1.24 moles NH3 of reacts with 1.79 moles of NO?

kotegsom [21]

Answer:

Option 1. NO

Explanation:

The balanced equation for the reaction is given below below:

4NH₃ + 6NO —> 5N₂ + 6H₂O

From the balanced equation above,

4 moles of NH₃ reacted with 6 moles of NO.

Finally, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

4 moles of NH₃ reacted with 6 moles of NO.

Therefore, 1.24 moles of NH₃ will react with = (1.24 × 6)/4 = 1.86 moles of NO

From the calculation made above, we can see that a higher amount of NO (i.e 1.86 moles) than what was given (i.e 1.79 moles) is needed to react completely with 1.24 moles of NH₃.

Therefore, NO is the limiting reactant and NH₃ is the excess reactant.

Thus, the 1st option gives the correct answer to the question

6 0

3 years ago

Read 2 more answers

Nitrogen and oxygen do not react appreciably at room temperature, as illustrated by our atmosphere. But at high temperatures, th

Gre4nikov [31]

Answer : The equilibrium concentration of NO is, 0.0092 M.

Solution :

First we have to calculate the concentration of NO.

$\text{Concentration of NO}=\frac{\text{Moles of }NO}{\text{Volume of solution}}=\frac{0.3152mol}{2.0L}=0.1576M$