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Radda [10]
3 years ago
10

What is the answer: 20°C to °F?

Chemistry
1 answer:
Masteriza [31]3 years ago
3 0

Answer:

68 degrees fahrenheit

Explanation:

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A balloon contains 0.950 mol of nitrogen gas and has a volume of 25.5 L. How many grams of N2 should be released from the balloo
kirill [66]
Answer:
Mass released = 8.6 g
Explanation:
Given data:
Initial number of moles nitrogen= 0.950 mol
Initial volume = 25.5 L
Final mass of nitrogen released = ?
Final volume = 17.3 L
Solution:
Formula:
V₁/n₁ = V₂/n₂
25.5 L / 0.950 mol = 17.3 L/n₂
n₂ = 17.3 L× 0.950 mol/25.5 L
n₂ = 16.435 L.mol /25.5 L
n₂ = 0.644 mol
Initial mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.950 mol × 28 g/mol
Mass = 26.6 g
Final mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.644 mol × 28 g/mol
Mass = 18.0 g
Mass released = initial mass - final mass
Mass released = 26.6 g - 18.0 g
Mass released = 8.6 g
3 0
3 years ago
True/False: The Northern Hemisphere experiences winter when Earth is farthest from the sun.
baherus [9]

Answer:

False

Explanation:

Apex

7 0
3 years ago
A ________________ is formed when two or more __________ bond by a covalent bond.
sergey [27]
Compound is formed when two or more atomic bond (or atom)
3 0
4 years ago
It is found that 250 ml of gas at stp has a mass of .675 g what is the molar mass
katrin2010 [14]

Answer:

61.3 g/mol  

Step-by-step explanation:

We can use the <em>Ideal Gas Law</em> to solve this problem:

       pV = nRT

Since n = m/M, the equation becomes

       pV = (m/M)RT     Multiply each side by M

    pVM = RT              Divide each side by RT

         M = (mRT)/(pV)

<em>Data: </em>

m = 0.675 g

R = 0.0.083 14 bar·L·K⁻¹mol⁻¹

T = 0 °C = 273.15 K

p = 1 bar

V = 250 mL = 0.250 L

<em>Calculation: </em>

M= (0.675 × 0.083 14 × 273.15)/(1 × 0.250)

M= 15.33/0.250

M= 61.3 g/mol

4 0
3 years ago
How do you calculate a wavelength of light emitted when
Novosadov [1.4K]

Answer: Wavelength of light emitted when  the electron in doubly ionized lithium makes a transition  from E12 to E8 is  0.1050\times 10^{-4m

Explanation:

Li:3:1s^22s^1

Li^{2+}:1:1s^1

Using Rydberg's Equation for hydrogen and hydrogen like atom:

\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )

Where,

\lambda = Wavelength of radiation

R_H = Rydberg's Constant

n_f = Higher energy level  

n_i= Lower energy level

We have:

n_f=8, n_i=12

R_H=1.097\times 10^7 m^{-1}

\frac{1}{\lambda}=1.097\times 10^7 m^{-1}\times \left(\frac{1}{12^2}-\frac{1}{8^2} \right )

\frac{1}{\lambda}=1.097\times 10^7 m^{-1}\times \frac{5}{576}

\frac{1}{\lambda}=9.523\times 10^4 m

\frac{1}{\lambda}=9.523\times 10^4 m

{\lambda}=0.1050\times 10^{-4}m

The wavelength of the photon emitted when the hydrogen atom undergoes a transition is 0.1050\times 10^{-4m

8 0
3 years ago
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