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joja [24]
3 years ago
8

If a chemist has a stock solution of HBr that is 10.0 M and would like to make 450.0 mL of 3.0 M HBr, how would he do it?

Chemistry
1 answer:
asambeis [7]3 years ago
4 0

This is a dilution that requires a certain volume from the stock solution to be diluted with distilled water to make a solution of HBr with a lesser concentration than the stock solution

Following dilution formula can be used

c1v1 = c2v2

Where c1 is concentration and v1 is the volume of the stock solution

c2 is concentration and v2 is volume of the diluted solution to be prepared

Substituting these values

10.0 M x v1 = 3.0 x 450.0 mL

v1 = 135.0 mL

A volume of 135.0 mL from HBr stock solution needs to be taken and diluted with distilled water upto 450.0 mL. The resulting solution will have a concentration of 3.0 M

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4 years ago
Which equation shows a conservation of mass?
pychu [463]
(4) PCl5--> PCl3+ Cl2 is the correct answer because we both have 1 P atom and 5 Cl atoms on both sides, and thus the masses of the elements don't change.
Hope this would help~
8 0
4 years ago
Determine the kinetic energy of an 833.0 kg roller coaster car moving at a speed of 20.0 m/s
trasher [3.6K]

Answer:

166,600J

Explanation:

Kinetic energy (K.E), which is the energy due to motion of a body, can be calculated by using the formula;

K.E = 1/2 × m × v²

Where;

K.E = kinetic energy (joules)

m = mass of body (kg)

v = speed or velocity (m/s)

According to this question, the mass of the roller coaster is 833.0 kg while its velocity/speed is 20.0m/s.

K.E = 1/2 × 833 × 20²

K.E = 1/2 × 833 × 400

K.E = 1/2 × 333200

K.E = 166,600

Therefore, the kinetic energy of the roller coaster car is 166,600J.

6 0
3 years ago
The state of matter for an object that has both definite volume and definite shape is
Romashka-Z-Leto [24]

The state of matter for an object that has both definite volume and definite shape is solid.

3 0
4 years ago
(C)
kari74 [83]

Answer:

177.3kg C₂₁H₄₄

Explanation:

Based on the chemical reaction:

C₂₁H₄₄ → 3C₂H₄ + C₁₅H₃₂

<em>Where 1 mole of C₂₁H₄₄ produce 3 moles of ethene, C₂H₄.</em>

<em />

To solve this question we need to determine the moles of ethene in 50.4kg. 1/3 these moles are the moles of C₂₁H₄₄ that must be added:

<em>Moles Ethene -Molar mass: 28.05g/mol-</em>

50.4kg = 50400g * (1mol / 28.05g) = 1796.8 moles of ethene

<em>Moles C₂₁H₄₄:</em>

1796.8 moles of ethene * (1 mol C₂₁H₄₄ / 3 mol C₂H₄) = 589.93 moles C₂₁H₄₄

<em>Mass C₂₁H₄₄:</em>

589.93 moles C₂₁H₄₄ * (296g / mol) = 177283g =

<h3>177.3kg C₂₁H₄₄</h3>
6 0
3 years ago
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