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USPshnik [31]
3 years ago
13

A sphere has a mass of 64 g, and a density of 2.0 g/cm3 . what is the radius of the sphere

Chemistry
1 answer:
slega [8]3 years ago
4 0
The size v=masse/density
v= 4*pi*R^3
R=(3*masse/(4*pi*density))^(1/3)
R=1.9695 cm
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Draw a line-bond structure for CBrN. Explicitly draw all H atoms. You do not have to include lone pairs in your answer. In cases
DochEvi [55]

Answer:

Br - C ≡ N

Explanation:

To draw the Lewis line-bond structure we need to bear in mind the octet rule, which states that in order to gain stability each <em>atom tends to share electrons until it has 8 electrons in its valence shell</em>.

  • C has 4 e⁻ in its valence shell so it will form 4 covalent bonds.
  • Br has 7 e⁻ in its valence shell so it will form 1 covalent bond.
  • N has 5 e⁻ in its valence shell so it will form 3 covalent bonds.

The most stable structure that respects these premises is:

Br - C ≡ N

It does not have any H atom.

8 0
3 years ago
A balloon has a volume of 6.2 liters at 23.2 C. The balloon is then heated to a temperature of 144.0 C. What is the volume of th
Marina CMI [18]

Answer:

8.7 L

Explanation:

T2(V1/T1) = V2

417.15 K(6.2 L/296.45 K) = 8.7 L

Remember to almost always change celcius to kelvin. Also, this is part of Charle's Law (temp and volume are proportional, so if temp increaces so must the volume or vice versa). Lastly, Charle's Law has the formula of V1/T1 = V2/T2. I just rearranged it to go along with your problem. Hence, the T2(V1/T1) = V2

4 0
4 years ago
What factors affect the dynamic state of equilibrium in a chemical reaction and how?
yanalaym [24]

Answer:

Only changes in temperature will influence the equilibrium constant K_c. The system will shift in response to certain external shocks. At the new equilibrium Q will still be equal to K_c, but the final concentrations will be different.

The question is asking for sources of the shocks that will influence the value of Q. For most reversible reactions:

  • External changes in the relative concentration of the products and reactants.

For some reversible reactions that involve gases:

  • Changes in pressure due to volume changes.

Catalysts do not influence the value of Q. See explanation.

Explanation:

\displaystyle K_c = {e}^{\Delta G/(R\cdot T)}.

Similar to the rate constant, the equilibrium constant K_c depends only on:

  • \Delta G the standard Gibbs energy change of the reaction, and
  • T the absolute temperature (in degrees Kelvins.)

The reversible reaction is in a dynamic equilibrium when the rate of the forward reaction is equal to the rate of the backward reaction. Reactants are constantly converted to products; products are constantly converted back to reactants. However, at equilibrium Q = K_c the two processes balance each other. The concentration of each species will stay the same.

Factors that alter the rate of one reaction more than the other will disrupt the equilibrium. These factors shall change the rate of successful collisions and hence the reaction rate.

  • Changes in concentration influence the number of particles per unit space.
  • Changes in temperature influence both the rate of collision and the percentage of particles with sufficient energy of reaction.

For reactions that involve gases,

  • Changing the volume of the container will change the concentration of gases and change the reaction rate.

However, there are cases where the number of gases particles on the reactant side and the product side are equal. Rates of the forward and backward reaction will change by the same extent. In such cases, there will not be a change in the final concentrations. Similarly, catalysts change the two rates by the same extent and will not change the final concentrations. Adding noble gases will also change the pressure. However, concentrations stay the same and the equilibrium position will not change.

8 0
3 years ago
Type of reaction<br> PbSO. →<br> PbSO3 + __02<br> Type of reaction:
Marysya12 [62]

Answer:

1.

work out the mean mode median and range

Explanation:

6 0
2 years ago
2. How many moles are in 2.8 Liters of CO2 gas?
andrew-mc [135]

Answer:

0.125 moles

Explanation:

2.8 litres is equivalent to 2.8dm³

At STP,

1 mole = 22.4 dm³

x mole = 2.8 dm³

Cross multiply

22.4x = 2.8

Divide both sides by 22.4

x = 2.8/22.4

x = 0.125

4 0
2 years ago
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