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enyata [817]
3 years ago
13

What type of radiation is carbon emitting in the following equation? 14/6C - 0/-1 + 14/7N A. alpha particles B. beta particles C

. gamma rays
Chemistry
2 answers:
dolphi86 [110]3 years ago
7 0
I think the answer is C
pickupchik [31]3 years ago
6 0

Answer is: B. beta particles.

Nuclear reaction: ¹⁴C → ¹⁴N + e⁻ + νe (electron antineutrino).

In beta minus decay (atomic number Z is increased by one, from 6 in carbon to 7 in nitrogen) neutron is converted to a proton and an electron and an electron antineutrino.

Beta decay is radioactive decay in which a beta ray and a neutrino are emitted from an atomic nucleus.

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When an alkaline earth metal, a, reacts with a halogen, x, the formula of the covalent compund formed should be a2x?
Ierofanga [76]

Alkaline earth metal are the elements present in II group in the periodic table and are known as 'Metals' and have a charge of +2.

Alkaline earth metals - Be , Mg Ca, Sr , Ba, Ra

Halogens are present in VII A group in the periodic table and are 'Non-metals' and have a charge of -1.

Halogens - F, Cl, Br, I, At

When Alkaline earth metal (metals) combine with Halogens (non-metals) the compound formed will be ionic compound and the formula of the compound will be based on the charges of the element.

When we write the formula of the ionic compound the charges of the elements get criss crossed.

For example - Mg (Alkaline earth metal) have a charge of +2 and Cl (Halogen) have a charge of -1 and when they combine to form the formula their charges get criss crossed and we will get Mg_{1}Cl_{2} or MgCl_{2}

When an alkaline earth metal, A, reacts with a halogen, X, the formula of the Ionic compound formed should be AX_{2}


3 0
3 years ago
Question 7
chubhunter [2.5K]
This answer is D or c
5 0
3 years ago
Draw an alkyl bromide with proper stereochemistry that can be used to synthesize the given alkene as the exclusive product via a
enot [183]

Answer:

See explanation below

Explanation:

You forgot to put the picture to do so. In this case, I manage to find one, and I hope is the one you are looking for. If not, then post it again and I'll gladly help you out again.

According to the picture with the answer, we have a cyclohexane with 4 methyl groups there. Two of them are facing towards the molecule with a darker bond. This means that the alkyl bromide, should have a bromine in one of the bonds, and in order to produce an E2 reaction, this bromine should be facing in the opposite direction of the methyl groups which are facing towards. This is because an E2 reaction occurs with the less steric hindrance in the molecule. If the bromine is in the same direction as the methyl group, it will cause a lot more of work to do a reaction, and therefore, an E2 reaction. I will promote instead a E1 or a sustitution product.

Therefore the alkyl bromide should be like the one in the picture 2.

3 0
3 years ago
If the mass of the object below is 28g, what is the density of the object below. Units are in cm below. Please round your answer
Murljashka [212]

Answer:

d = 0.93 g/cm³

Explanation:

Given data:

Mass of object = 28 g

Volume of object = 3cm×2cm×5cm

density of object = ?

Solution:

Volume of object = 3cm × 2cm ×5cm

Volume of object = 30 cm³

Density of object:

d = m/v

by putting values,

d = 28 g/ 30 cm³

d = 0.93 g/cm³

3 0
3 years ago
When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4 g of carbon were burned in the presence of
Natasha2012 [34]

When carbon reacts with oxygen it forms CO2. This can depicted by the below equation.

C + O2→ CO2

It has been mentioned that when 14.4 g of C reacts with 53.9 g of O2, then 15.5 g of O2 remains unreacted. <u>This indicates that Carbon is the limiting reagent and hence the amount of CO2 produced is based on the amount of Carbon burnt.</u>

C + O2→ CO2

In the above equation , 1 mole of carbon reacts with 1 mole of O2 to produce 1 mole of CO2.

In this case 14.4 g of Carbon reacts with 53.9 of O2 to produce "x"g of CO2.

<u>No of moles = mass of the substance÷molar mass of the substance</u>

No of moles of carbon = 14.4 /12= 1.2 moles

No of moles of O2 = Mass of reacted O2/Molar mass of O2.

No of moles of O2 = (Total mass of O2 burned - Mass of unreacted O2)/32

No of moles of O2 = (53.9-15.5) ÷ 32 = 1.2 moles.

Hence as already discussed 1 mole of Carbon reacts with 1 mole of O2 to produce 1 mole of CO2. In this case 1.2 moles of carbon reacts with 1.2 moles of O2 to produce 1.2 moles of CO2.

Moles of carbon dioxide = Mass of CO2 produced /Molar mass of CO2

Mass of CO2 produced(x) = Moles of CO2 ×Molar mass of CO2

Mass of CO2 produced(x) = 1.2 x 44 = 52.8 g

<u>Thus 52.8 g of CO2 is produced.</u>

5 0
3 years ago
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