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evablogger [386]
3 years ago
5

Select the correct answer.

Chemistry
1 answer:
Artist 52 [7]3 years ago
3 0
I think c cause if he is in a relationship that’s abusive he needs to speak up before it goes way worst.
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An FAS calibration curve (absorbance, y-axis vs. FAS in g/mL) gave an equation he regression line of: y-3678(x)+0.056. If an unk
mezya [45]

Answer:

FAS concentration = 1.61*10^-4M

Explanation:

Beer Lambert's law relates the absorbance (A) of a substance to its concentration (c) as:

A = \epsilon lc----(1)

where ε = molar absorption coefficient

l = path length

A plot of 'A' vs 'c' gives a straight line with slope = εl

In addition absorbance (A) is related to % Transmittance (%T) as:

A = 2-log%T----(2)

For the FAS solution, the corresponding calibration fit is given as:

y = 3678(x) + 0.056

This implies that the slope = εl = 3678

It is given that %T = 25.6%

A = 2-log(25.6)=0.592

Based on equation(1):

c = \frac{A}{\epsilon l}=\frac{0.592}{3678}=1.61*10^{-4}M

8 0
2 years ago
Can someone please help me?
irina [24]
Answer:
your answer is hawks
3 0
2 years ago
Read 2 more answers
Which is an example of a solution?
Rudiy27
The answer is C. Salt and water is a solution 
3 0
3 years ago
Read 2 more answers
The volume of a sample gas, initially at 25 C and 158 mL, increased to 450 mL. What is the final temperature of the sample of ga
Rashid [163]

Answer:

Final temperature of the gas is  576 ^{0}\textrm{C}.

Explanation:

As the amount of gas and pressure of the gas remains constant therefore in accordance with Charles's law:

                                       \frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}

where V_{1} and V_{2} are volume of gas at T_{1} and T_{2} temperature (in kelvin scale) respectively.

Here V_{1}=158mL , T_{1}=(273+25)K=298K and V_{2}=450mL

So  T_{2}=\frac{V_{2}T_{1}}{V_{1}}=\frac{(450mL)\times (298K)}{(158mL)}=849K 

849 K = (849-273) ^{0}\textrm{C} = 576 ^{0}\textrm{C}

So final temperature of the gas is  576 ^{0}\textrm{C}.

3 0
3 years ago
Zinc oxide adopts a face-centered cubic arrangement. Both Zinc ions and oxide ions adopt the face-centered cubic arrangement; wi
vichka [17]

Answer:

5.41 ×10⁻²²

Explanation:

We were told right from the question that both the Zinc ions and the Zinc oxide adopts a face-centered cubic arrangement.

Then, the number of ZnO molecule in one unit cell = 4

The standard molar mass of ZnO = 81.38g

Avogadro's constant = 6.023 × 10²³ mole

∴

The mass of one unit cell of zinc oxide can be calculated as:

= \frac{4*81.38}{6.023*10^{23}}

= 5.40461564×10⁻²²

≅ 5.41 ×10⁻²²

∴ The mass of one unit cell of zinc oxide = 5.41 ×10⁻²²

3 0
2 years ago
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