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musickatia [10]
3 years ago
13

Plz Help Me

Chemistry
2 answers:
telo118 [61]3 years ago
8 0

The answer is D.) 1.36 x 1012 m

Rasek [7]3 years ago
7 0

Answer:- The wavelength is 1.36*10^1^2m and the right option is D.

Solution:-  Wavelength is inversely proportional to the frequency and the equation is:

\lambda =\frac{c}{\nu }

where, \lambda
 is the wavelength, c is the speed of light and \nu is frequency.

Frequency is 2.20*10^-^4Hz and the speed of light is 3.00*10^8\frac{m}{s} .

Since, 1Hz=1s^-^1

So, 2.20*10^-^4Hz=2.20*10^-^4s^-^1

Let's plug in the values in the equation:

\lambda =\frac{3.00*10^8m.s^-^1}{2.20*10^-^4s^-^1}

\lambda =1.36*10^1^2m

Hence, the right option is D) 1.36*10^1^2m .





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Alex_Xolod [135]

Answer:

74 g/mol

Explanation:

Using a periodic table, we can determine the molar mass by adding together 1 Ca, 2 O, and 2 H. This turns out to be approximately (40+32+2) = 74 g/mol

3 0
3 years ago
The value of the equilibrium constant K depends on: I. the initial concentrations of the reactants. II. the initial concentratio
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Answer:

None of these

Explanation:

For a reaction;

aA + bB ------>cC + dD

The equilibrium constant K is given as;

K = [C]^c [D]^d/[A]^a [B]^b

The equilibrium constant neither depends on the concentrations of the reactants nor on that of the products.

Let us recall that at equilibrium, the concentrations of reactants and products remain largely constant. This implies that, concentration of species do not appreciably change at equilibrium because the rates of forward and reverse reactions are equal.

Hence, the equilibrium constant neither depends on the initial/final concentrations of the reactants nor on the initial/final concentrations of the products.

4 0
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The vapor pressure of ethanol is 400 mmhg at 63.5°c. its molar heat of vaporization is 39.3 kj/mol. what is vapor pressure of et
SVETLANKA909090 [29]

Answer:- The pressure of ethanol would be 109 mmHg.

Solution:- This problem is based on Clausius clapeyron equation--

ln(\frac{P_1}{P_2})=(\frac{\Delta Hvap}{R})(\frac{1}{T_2}-\frac{1}{T_1})

Given, T_1 = 63.5 + 273 = 336.5 K

T_2 = 34.9 + 273 = 307.9 K

P_1 = 400 mmHg

P_2 = ?

\Delta Hvap = 39.3 kJ/mol = 39300 J/mol

R = 8.314 J/mol.K

Let's plug in the values in the equation and do the calculations.

ln(\frac{400}{P_2})=(\frac{39300}{8.314})(\frac{1}{307.9}-\frac{1}{336.5})

ln(\frac{400}{P_2}) = 1.30

On taking anti ln to both sides...

\frac{400}{P_2} = e^1^.^3^0

\frac{400}{P_2} = 3.67

P_2 = 400/3.67

P_2 = 109 mmHg

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The process by which an animal becomes accustomed to its situation is referred to as A. Habituation B nurturing C aggression D d
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For the following reaction, 8.70 grams of benzene (C6H6) are allowed to react with 13.7 grams of oxygen gas. benzene (C6H6) (l)
artcher [175]

Answer:

Maximum amount of carbon dioxide that can be formed → 7.52 g

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We need to determine the limting reactant so we convert the mass to moles:

8.70 g. 1mol / 78g = 0.111 moles of benzene

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15 moles of O₂ can produce 6 moles of CO₂

So, 0.428 moles of O₂ may produce (0.428 . 6)/ 15 = 0.171 moles of CO₂

We convert the moles to mass → 0.171 mol . 44g / 1mol = 7.52 g

This is the maximum amount of carbon dioxide that can be formed

We convert the mass of the limiting reactant that remains after the reaction is complete → 0.404 mol . 32g / 1mol = 12.9 g of O₂

7 0
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