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VikaD [51]
3 years ago
11

Which atmospheric gas protects living things on Earth from gamma rays

Chemistry
2 answers:
Kamila [148]3 years ago
8 0
I think the answer is B.
I hope this helps u.(And I hope it's correct)!!!!!!!!!!!
slega [8]3 years ago
4 0
I think the answer is  B . I really hope that this helps you out.
You might be interested in
Name each subatomic particle, its charge, and its location in an atom.
Tpy6a [65]
Protons: charge +1, have a mass of 1 and are found in the nucleus 

Neutrons: charge 0, have a mass of 1 and are found in the nucleus

Electrons: charge -1, have a mass of 1/840 and are found on the outside of the nucleus 

hope that helps 
4 0
3 years ago
What is the minimum cation-to-anion radius ratio (rC/rA) for an octahedral interstitial lattice site?
mash [69]

Answer:

0. 414

Explanation:

Octahedral interstitial lattice sites.

Octahedral interstitial lattice sites are in a plane parallel to the base plane between two compact planes and project to the center of an elementary triangle of the base plane.

The octahedral sites are located halfway between the two planes. They are vertical to the locations of the spheres of a possible plane. There are, therefore, as many octahedral sites as there are atoms in a compact network.

The Octahedral interstitial void ratio range is 0.414 to 0.732. Thus, the minimum cation-to-anion radius ratio for an octahedral interstitial lattice site is 0. 414.

3 0
3 years ago
How does a liquid substance change into gas? ​
Afina-wow [57]

Answer: heat

Explanation:

Is that heat boils the liquid and the liquid eventually evaporated into gas

3 0
3 years ago
A penny has a mass of 2.50g and the Moon has a mass of ×7.351022kg, How many moles of pennies have a mass equal to the mass of t
baherus [9]
Answer: 122 moles

Procedure:

1) Convert all the units to the same unit

2) mass of a penny = 2.50 g

3) mass of the Moon = 7.35 * 10^22 kg (I had to arrage your numbers because it was wrong).

=> 7.35 * 10^22 kg * 1000 g / kg = 7.35 * 10^ 25 g.

4) find how many times the mass of a penny is contained in the mass of the Moon.

You have to divide the mass of the Moon by the mass of a penny

7.35 * 10^ 25 g / 2.50 g = 2.94 * 10^25 pennies

That means that 2.94 * 10^ 25 pennies have the mass of the Moon, which you can check by mulitiplying the mass of one penny times the number ob pennies: 2.50 g * 2.94 * 10^25 = 7.35 * 10^25.

5) Convert the number of pennies into mole unit. That is using Avogadros's number: 6.022 * 10^ 23

7.35 * 10^ 25 penny * 1 mol / (6.022 * 10^ 23 penny) = 1.22* 10^ 2 mole = 122 mol.

Answer: 122 mol
8 0
3 years ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
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