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VMariaS [17]
3 years ago
13

When the forward and reverse paths of a change occur at the same rate,

Chemistry
2 answers:
Zarrin [17]3 years ago
8 0
The correct answer is option B. When the forward and reverse paths of a change occur at the same rate, <span>the system is in equilibrium specifically in dynamic equilibrium.</span> Dynamic equilibrium is the balance in a process that is continuing. It is achieved in a reaction when the forward rate of reaction and the backward rate of reaction is at the same value or equal.

Norma-Jean [14]3 years ago
6 0

Answer:

b. the system is in equilibrium.

Explanation:

Chemical equilibrium occurs when any reaction is reversible, in which there are two possible reactions, one forward (where reactants turn into products) and a reverse one (where products turn reactants). These reactions have the same velocity.

In this state, the rates of forward and reverse reactions are equal, and concentrations of reagents and products remain unchanged.

The velocities within a reversible reaction (in equilibrium) can be represented as follows:

A + B ⇄ C + D

However, this state of equilibrium is vulnerable to some factors, such as changes in pressure, temperature and addition or withdrawal of one or more substances from the system.

These may change one of these velocities and consequently, modify the concentrations of reagents and/or products, causing an equilibrium shift.

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What are different type of plastids explain them?​
BARSIC [14]

Answer:

Explanation:

There are 3 types of plastids :-

1) Chloroplasts:- The green plastids which contain chlorophyll pigments for photosynthesis.

2) Chromoplasts:-The coloured plastids for pigment synthesis and storage.

3) Leucoplasts:- The colourless plastids for monoterpene synthesis found in non- photosynthetic parts of the plants.

They are of three types:-  

a) Amyloplasts- stores starch.

b) Proteinoplasts-  stores proteins.

c) Elaioplasts- stores fats and oils.

6 0
3 years ago
Consider this reaction: NH + + HPO 4 + NH3 + H2PO4
Serggg [28]

Answer:NH3

Explanation:Ecd

5 0
3 years ago
How many half-lives will pass by the time 1.56% of I-131 is present? B. Approximately how many days does that equal? *
serg [7]

Answer: Hmmmmm that's crazy....

There are a couple of equations one could use for this type of problem, but I find the following to be the easiest to use and to understand.

Fraction remaining (FR) = 0.5n

n = number of half lives that have elapsed

In this problem, we need to find n and are given the FR, which is 1.56% or 0.0156 (as a fraction).

0.0156 = 0.5n

log 0.0156 = n log 0.5

-1.81 = -0.301 n

n = 6.0 half lives have elapsed

Explanation:

Just wanted to help. Hopefully it's correct wouldn't want to waster your time ;)

6 0
2 years ago
A lithium flame has a characteristic red color due to emissions of wavelength 671 nm. What is the mass equivalence of 1 mol of p
e-lub [12.9K]

Answer:

1.98x10⁻¹² kg

Explanation:

The <em>energy of a photon</em> is given by:

  • E= hc/λ

h is Planck's constant, 6.626x10⁻³⁴ J·s

c is the speed of light, 3x10⁸ m/s

and λ is the wavelenght, 671 nm (or 6.71x10⁻⁷m)

  • E = 6.626x10⁻³⁴ J·s * 3x10⁸ m/s ÷ 6.71x10⁻⁷m = 2.96x10⁻¹⁹ J

Now we multiply that value by <em>Avogadro's number</em>, to <u>calculate the energy of 1 mol of such protons</u>:

  • 1 mol =  6.023x10²³ photons
  • 2.96x10⁻¹⁹ J *  6.023x10²³ = 1.78x10⁵ J

Finally we <u>calculate the mass equivalence</u> using the equation:

  • E=m*c²
  • m=E/c²
  • m =  1.78x10⁵ J / (3x10⁸ m/s)² = 1.98x10⁻¹² kg

6 0
3 years ago
How many grams of copper (II) nitrate would be produced from 0.80 g of copper metal reacting with excess nitric acid?
zaharov [31]

Answer:

m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2

Explanation:

Hello!

In this case, since the chemical reaction between copper and nitric acid is:

2HNO_3+Cu\rightarrow Cu(NO_3)_2+H_2

By starting with 0.80 g of copper metal (molar mass = 63.54 g/mol) and considering the 1:1 mole ratio between copper and copper (II) nitrate (molar mass = 187.56 g/mol) we can compute that mass via stoichiometry as shown below:

m_{Cu(NO_3)_2}=0.80gCu*\frac{1molCu}{63.54gCu} *\frac{1molCu(NO_3)_2}{1molCu} *\frac{187.56gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2

However, the real reaction between copper and nitric acid releases nitrogen oxide, yet it does not modify the calculations since the 1:1 mole ratio is still there:

4HNO_3+Cu\rightarrow Cu(NO_3)_2+2H_2O+2NO_2

Best regards!

7 0
3 years ago
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