If motion starts and stops at the same place is this distance zero ?

1 answer:

8 0

Displacement is zero. Displacement is in reference to the starting and ending positions, so starting and ending in the same place would cause a net displacement of zero. Distance, however, is not zero. It is a measurement of the total amount moved, so if the object left the starting position, the distance is automatically not zero.

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How many moles of nitric acid are present in 35.0 ml of a 2.20 M solution?

vodomira [7]

Answer:

There are 77 millimoles of nitric acid present in 35.0 mL of a 2.20 M solution

Explanation:

Molarity of the solution = 2.20 M

$Molarity=\frac{number\:of\:moles}{Volume\:of\:Solution\:in\:L}\\\\Number\:of\:moles=Molarity\times(Volume\:of\:Solution\:in\:L)\\\\Volume\:of\:Solution=35\:mL=35\times10^{-3}L\\\\Number\:of\:moles=2.20\times35\times10^{-3}=77\times10^{-3}\:moles\:of\:HNO_{3}$

Therefore, there are 77 millimoles of nitric acid present in 35.0 mL of a 2.20 M solution

8 0

3 years ago

How many grams are in 8.2 x 1022 molecules of N216?

hoa [83]

Answer:

N2I6 = 789 g

N2I6 = 8.2x1022 molecules N2I6 x 1 mole/6.02x1023 molecules = 1.36x10-1 moles = 0.136 moles

N2I6=0.136molesx789g/mole=107g=110g

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Hope this helps, have a BLESSED AND WONDERFUL DAY!

- Cutiepatutie ☺❀❤

7 0

3 years ago

When is friction advantageous and when is it not advantageous? Explain.

Novosadov [1.4K]

Answer:

here

Explanation:

FRICTION ADVANTAGES:

a) the soles of are shoes are designed in a way that they create friction and prevent us from slipping and falling.

b) We are capable of writing with a pen due to the friction between the pen and the paper.

FRICTION DISADVANTAGES:

b) Forest fires are caused due to the friction between the trees branches

( please vote me for brainliest )

5 0

3 years ago

When the equation Sn + HNO₃ → SnO₂ + NO₂ + H₂O is balanced in acidic solution, what is the smallest whole-number coefficient for

balu736 [363]

Answer:

Explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

$Sn^0 + H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 + N^{4+}O^{2-}_2 + H^+_2O^-$

Whereas the half reactions are:

$Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O$

Next, we exchange the transferred electrons:

$1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow 4N^{4+}O^{2-}_2+4H_2O$