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BARSIC [14]
3 years ago
15

A balloon moves vertically upward at

Chemistry
1 answer:
Usimov [2.4K]3 years ago
6 0

Answer:

B. Velocity

Explanation:

The answer is Velocity because the def of velocity is the rate of constant speed in a given direction. I hope this helps, have a good night.

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Identify the substances that will appear in the equilibrium constant expression for the equation: 2Ag+(aq)+Zn(s)<->Zn2+(aq
Simora [160]

Hey there!


We Know that:



 2 Ag⁺(aq) + Zn(s) <-> Zn²⁺(aq)+2 Ag(s)


The equilibrium expression for the reaction is:



Kc =  [ Zn⁺² ]  /  [Ag⁺ ]²


Hope that helps!

8 0
2 years ago
Read 2 more answers
2. A 10 kg ball is rolled down a rail from a height of 20 m. What is the velocity at the bottom of the
solmaris [256]

Answer:

19.8m/s

Explanation:

Given parameters:

Mass of the ball  = 10kg

Height of the rail  = 20m

Unknown:

Velocity at the bottom of the rail  = ?

Solution:

The velocity at the bottom of the rail is its final velocity.

Using the appropriate motion equation, we can find this parameter;

   V²   = U²  + 2gH

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

H is the height

  the ball was rolled from rest, U  = 0

  V²  = O²  + 2 x 9.8 x 20

 V  = 19.8m/s

3 0
2 years ago
Consider these changes.
Ede4ka [16]

Answer:

(a) The system does work on the surroundings.

(b) The surroundings do work on the system.

(c) The system does work on the surroundings.

(d) No work is done.

Explanation:

The work (W) done in a chemical reaction can be calculated using the following expression:

W = -R.T.Δn(g)

where,

R is the ideal gas constant

T is the absolute temperature

Δn(g) is the difference between the gaseous moles of products and the gaseous moles of reactants

R and T are always positive.

  • If Δn(g) > 0, W < 0, which means that the system does work on the surroundings.
  • If Δn(g) < 0, W > 0, which means that the surroundings do work on the system.
  • If Δn(g) = 0, W = 0, which means that no work is done.

<em>(a) Hg(l) ⇒ Hg(g)</em>

Δn(g) = 1 - 0 = 1. W < 0. The system does work on the surroundings.

<em>(b) 3 O₂(g) ⇒ 2 O₃(g) </em>

Δn(g) = 2 - 3 = -1. W > 0. The surroundings do work on the system.

<em>(c) CuSO₄.5H₂O(s) ⇒ CuSO₄(s) + 5H₅O(g) </em>

Δn(g) = 5 - 0 = 5. W < 0. The system does work on the surroundings.

<em>(d) H₂(g) + F₂(g) ⇒ 2 HF(g)</em>

Δn(g) = 2 - 2 = 0. W = 0. No work is done.

3 0
3 years ago
HELP PLEASE I GIVE BRAINLIST heeelllpp
kotykmax [81]
Here we go, 4 attachments, you made me work LOL

8 0
2 years ago
The dilution factor D for an unseeded mixture of wastewater is 0.05. The DO of the mixture is initially 8.0 mg/L, and after 5 da
enyata [817]

Answer:

(i) 5-day BOD of the waste is 120 mg/l.

(ii) The ultimate carbonaceous BOD (Lo) is 20 mg/l.

Explanation:

The dilution factor D is 0.05.

The initial DO is 8.0 mg/L and the DO after 5 days is 2.0 mg/L.

The BOD of the waste for an unseeded mixture is

BOD_5=(DO_5-DO_0)/D=(8-2)/0.05=6/0.05=120mg/l

The ultimate carbonaceous BOD (Lo) can be calculated as

L = L_o*10 ^{- k_1t} \\L_o=L/10^{- k_1t}=2/10^{- 0.2*5}=2/10^{- 1}=2*10=20mg/l

6 0
2 years ago
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