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ahrayia [7]
3 years ago
14

Which temperature equation do you use when trying to find Celsius?

Chemistry
1 answer:
Free_Kalibri [48]3 years ago
8 0

Answer: First subtract 32 degrees, then multiply the result by 5/9 (0.556 on a calculator). For example, suppose a thermometer reads 68 degrees Fahrenheit. Subtract 32 from 68, leaving 36. Multiply 36 by 5/9, which equals 20 degrees Celsius.

Explanation:

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What is the mass percent when 144 grams of NaCl is dissolved in 255 grams of water?
Daniel [21]

Answer:

36.09% approximately

Explanation:

I'm not good at English so I don't know if I get it right

4 0
3 years ago
What volume in milliliters of 0.100 M HCIO3 is required to neutralize<br> 40.0 mL of 0.165 M KOH?
Veronika [31]

Answer:

66.0 mL HClO3

Explanation:

M1V1 = M2V2

M1 = 0.100 M HClO3

V1 = ?

M2 = 0.165 M KOH

V2 = 40.0 mL KOH

Solve for V1 --> V1 = M2V2/M1

V1 = (0.165 M)(40.0 mL) / (0.100 M) = 66.0 mL HClO3

7 0
3 years ago
Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:
ololo11 [35]

Hence the order of the transition in order of increasing frequency of the photon absorbed or emitted  will be : d < a < c < b

E = -13.6×Z²/n²

where,

E = energy of  orbit

n = number of orbit

Z = atomic number

Energy of n = 1 in an hydrogen atom:

E₁ = -13.6× 1²/1² = -13.6eV

Energy of n = 2 in an hydrogen atom:

E₂ = -13.6 × 1²/2² = -3.40eV

Energy of n = 3 in an hydrogen atom:

E₃ = -13.6× 1²/3² = -1.51eV

Energy of n = 4 in an hydrogen atom:

E₄ = -13.6× 1²/4² = -0.85eV

Energy of n = 5 in an hydrogen atom:

E₃ = -13.6× 1²/ 5² = -0.54eV

a) n = 2 to n = 4 (absorption)

ΔE₁ = E₄ - E₂ = -0.85- (- 3.40) = 2.55eV

b) n = 2 to n = 1 (emission)

Δ E₂ = E₁ - E₂ = -13.6 - (- 3.40) = -10.2eV

Negative sign indicates that emission will take place.

c) n = 2 to n = 5 (absorption)

ΔE₃ = E₅ - E₂ = -0.54 -( -3.40) = -2.85eV

d) n = 4 to n = 3 (emission)

ΔE₄ = E₃ - E₄ = -1.51 - (- 0.85) = - 0.66eV

Negative sign indicates that emission will take place.

According to Planck's equation, higher the frequency of the wave higher will be the energy:

E = hv

h = Planck's constant

v =frequency of the wave

So, the increasing order of magnitude of the energy difference :

E₄< E₁ <E₃ <E₂

The  H atom electron transitions in order of increasing frequency of the photon absorbed or emitted will be d < a < c < b

: d < a < c < b

To learn more about transitions visit the link:

brainly.com/question/28304182?referrer=searchResults

The question is incomplete , complete question is:

Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:

(a) n = 2 to n = 4

(b) n = 2 to n = 1

(c) n = 2 to n = 5

(d) n = 4 to n = 3

#SPJ4

6 0
2 years ago
Normality (N) of concentrated<br>H2SO4 is<br>a. 12<br>b. 36<br>c. 24<br>d. 48<br>​
kolezko [41]

Answer:

N=100×10÷49= 20.41 near Answer is point C

6 0
4 years ago
HELP PLEASE <br> look at attached pictures <br> Part B
shusha [124]

Answer:

what do you need help with

Explanation:

7 0
3 years ago
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