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sashaice [31]
3 years ago
7

What is the frequency of an x-ray with a wavelength of 1.15 times 10^-10?

Chemistry
1 answer:
diamong [38]3 years ago
5 0
The answer to the problem is 11.615
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Which block contains 5 orbitals?
exis [7]
D sublevel because the s sublevel has one orbital, the p sublevel has three orbitals, the d sublevel has five orbitals, and the f sublevel has seven orbitals. In the first period, only the 1s sublevel is being filled.
4 0
2 years ago
Confused as heck. please help!
Nitella [24]

Answer: that all thre water cycle and C is vaporation

Explanation:

6 0
2 years ago
Tetrachloromethane, CCl4 is produced from the substitution reaction between methane and chlorine gas. If the rate of formation o
Korolek [52]

The rate of disappearance of chlorine gas : 0.2 mol/dm³

<h3>Further explanation</h3>

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.

For reaction :

\tt aA+bB\rightarrow cC+dD

The rate reaction :

\tt -\dfrac{1}{a}\dfrac{d[-A]}{dt}= -\dfrac{1}{b}\dfrac{d[-B]}{dt}=\dfrac{1}{c}\dfrac{d[C]}{dt}=\dfrac{1}{d}\dfrac{d[D]}{dt}

Reaction for formation CCl₄ :

<em>CH₄+4Cl₂⇒CCl₄+4HCl</em>

<em />

From equation, rate of reaction = rate of formation CCl₄ = 0.05 mol/dm³

Rate of formation of  CCl₄  = reaction rate x coefficient of  CCCl₄

0.05 mol/dm³ = reaction rate x 1⇒reaction rate = 0.05 mol/dm³

The rate of disappearance of chlorine gas (Cl₂) :

Rate of disappearance of  Cl₂  = reaction rate x coefficient of  Cl₂

Rate of disappearance of  Cl₂ = 0.05 x 4 = 0.2 mol/dm³

4 0
3 years ago
An empty beaker is weighed and found to weigh 23.1 g. Some potassium chloride is then added to the beaker and weighed again. The
GuDViN [60]

Answer:Mass of Potassium chloride =1.762g

Explanation:

Mass of empty beaker = 23.100 g

Mass of beaker with Potassium chloride = 24.862g

Mass of Potassium chloride = Final weight - initial weight = Mass of beaker with Potassium chloride  - Mass of empty beaker = 24.862-23.100 = 1.762g

8 0
3 years ago
Determine the work done by an ideal gas while expanding by a volume of 0.25 L against an external pressure of 1.50 atm (assume a
Sliva [168]

Answer : The value of work done by an ideal gas is, 37.9 J

Explanation :

Formula used :

Expansion work = External pressure of gas × Volume  of gas

Expansion work = 1.50 atm × 0.25 L

Expansion work = 0.375 L.atm

Conversion used : (1 L.atm = 101.3 J)

Expansion work = 0.375 × 101.3 = 37.9 J

Therefore, the value of work done by an ideal gas is, 37.9 J

6 0
3 years ago
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