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Zigmanuir [339]
3 years ago
10

Radon-219 decays to radon-218 by releasing... a. a positron b.a muon O c. a neutron O d. an electron O e. a proton

Chemistry
1 answer:
MrRa [10]3 years ago
4 0

Answer: Option (c) is the correct answer.

Explanation:

A positron is denoted as ^{0}_{+1}e, a neutron is denoted as ^{1}_{0}n, an electron is denoted as ^{0}_{-1}e, and a proton is denoted as ^{1}_{1}p.

Therefore, when ^{219}Rn will decay into ^{218}Rn this means there occurs decrease in its mass number. Hence, it means a neutron will be releasing during this reaction.

The reaction will be as follows.

                ^{219}Rn \rightarrow ^{218}Rn + ^{1}_{0}n

You might be interested in
Copper has a density of 8.96 g/cm3. If 75.0 g of copper is added to 50.0 mL of water in a graduated cylinder, to what volume rea
Tcecarenko [31]

Answer:

The answer to your question is    Final volume = 58.37 ml

Explanation:

Data

density = 8.96 g/cm³

mass = 75 g

volume of water = 50 ml

Process

1.- Calculate the volume of copper

  Density = mass / volume

Solve for volume

  Volume = mass / density

Substitution

  Volume = 75/8.96

Simplification

  Volume = 8.37cm³    or 8.37 cm³

2.- Calculate the new volume of water in the graduated cylinder

  Final volume = 50 + 8.37

  Final volume = 58.37 ml

3 0
3 years ago
What is the order of increasing rate of eusion for the following gases, Ar, CO2, H2, N2?
AVprozaik [17]

Answer:

H2 > N2 > Ar > CO2

Explanation:

Graham's law explains why some gases efuse faster than others. This is due to the difference i their molar mass. Generally; The rate of effusion of gaseous substances is inversely proportional to the square rot of its molar mass.

This means gases with low molar masses would have higher efusion rate compared to gases with higher molar masses.

So now we just need to compare the molar masses of the various gases;

Ar - 39.95

CO2 - 44.01

H2 - 2

N2 - 28.01

To obtain the order in increasing rate, we have to order the gases in decreasing molar mass. This order of increasing rate is given as;

H2 > N2 > Ar > CO2

8 0
3 years ago
A spectrophotometer measures the transmittance or the absorbance, or both, of a particular wavelength of light after it has pass
Cerrena [4.2K]

Answer:Answer: The step that is NOT necessary to complete before a cuvette is placed into the spectrophotometer is option B (Write, in ink, either sample or blank on the side of the cuvette to keep track of them)

Explanation: spectrophotometer is an instrument used to measure the light intensity absorbed after being passed through a solution. Before the absorbance of the sample solution, a solvent solution called blank is used for the calibration of the machine and this blank solvent is placed in a cuvette. The procedure usually comes first before the main sample is processed. Therefore there is no need to

Write, in ink, either sample or blank on the side of the cuvette to keep track of them. This is so since sample and blank is not absorbed at the same time by the machine.

7 0
4 years ago
Read 2 more answers
A sample was collected from a computer keyboard and a toilet seat using two separate cotton swabs. Then, each swab was wiped on
Gnom [1K]

Answer:

toilet seat and washing of hands ✋ very well

5 0
3 years ago
The constant
Zigmanuir [339]

Answer:

75 kJ/mol

Explanation:

The reactions occur at a rate, which means that the concentration of the reagents decays at a time. The rate law is a function of the concentrations and of the rate constant (k) which depends on the temperature of the reaction.

The activation energy (Ea) is the minimum energy that the reagents must have so the reaction will happen. The rate constant is related to the activation energy by the Arrhenius equation:

ln(k) = ln(A) -Ea/RT

Where A is a constant of the reaction, which doesn't depend on the temperature, R is the gas constant (8.314 J/mol.K), and T is the temperature. So, for two different temperatures, if we make the difference between the two equations:

ln(k1) - ln(k2) = ln(A) - Ea/RT1 - ln(A) + Ea/RT2

ln (k1/k2) = (Ea/R)*(1/T2 - 1/T1)

k1 = 8.3x10⁸, T1 = 142.0°C = 415 K

k2 = 6.9x10⁶, T2 = 67.0°C = 340 K

ln(8.3x10⁸/6.9x10⁶) = (Ea/8.314)*(1/340 - 1/415)

4.8 = 6.39x10⁻⁵Ea

Ea = 75078 J/mol

Ea = 75 kJ/mol

4 0
3 years ago
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