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Llana [10]
3 years ago
10

Element used to make first atomic bomb

Chemistry
1 answer:
muminat3 years ago
5 0
The element used to make the first atomic bomb is uranium
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What is the general structure for a chemical equation?
kramer
<span>A </span>chemical formula<span> is a way of expressing information about the proportions of </span>atoms<span> that constitute a particular</span>chemical compound<span>, using a single line of </span>chemical element<span> symbols, numbers, and sometimes also other symbols, such as parentheses, dashes, brackets, commas and </span>plus<span> (+) and </span>minus<span> (−) signs. A chemical formula is not a </span>chemical name showing how the atoms are arranged. 
6 0
3 years ago
Draw a Lewis structure for SO 2 in which all atoms obey the octet rule. Show formal charges. Do not consider ringed structures.
blagie [28]

Answer:

The answer is attached below

Explanation:

To draw the Lewis structure the sulphur willl be placed in the centre with the Valence electron sorrounding it the electrons between sulphu and oxygen to form bonds

5 0
3 years ago
In the equation CH4 + 2O2 --&gt; 2H2O + CO2 What is the mass of CO2 produced when 35g of O2 reacts?
Anestetic [448]

Answer:

24.06 g of CO₂

Explanation:

The balanced equation for the reaction is given below:

CH₄ + 2O₂ —> 2H₂O + CO₂

Next, we shall determine the mass of O₂ that reacted and the mass of CO₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mass of O₂ from the balanced equation = 2 × 32 = 64 g

Molar mass of CO₂ = 12 + (2×16)

= 12 + 32

= 44 g/mol

Mass of CO₂ from the balanced equation = 1 × 44 = 44 g

SUMMARY:

From the balanced equation above,

64 g of O₂ reacted to produce 44 g of CO₂.

Finally, we shall determine the mass of CO₂ produced by the reaction of 35 g of O₂. This can be obtained as follow:

From the balanced equation above,

64 g of O₂ reacted to produce 44 g of CO₂.

Therefore, 35 g of O₂ will react to produce = (35 × 44)/64 = 24.06 g of CO₂.

Thus, 24.06 g of CO₂ were produced from the reaction.

8 0
3 years ago
Modify the given fatty acid so that it represents the 18‑carbon fatty acid designated 18:2(Δ9,12). Draw any double bonds in the
Mekhanik [1.2K]

Answer:

See explanation

Explanation:

In this case, we have to remember the meaning of the nomenclature "18:2Δ9,12".  Where 18 is the <u>number of carbon atom</u>s, 2 is the <u>number of double bonds,</u> and the numbers successive to Δ "delta" the position of the double bonds <u>starting</u> to count from the carboxylic -COOH end of the molecule.

In other words, the main functional group is a <u>carboxylic acid</u>. We have a total of 18 carbons. Additionally, we have 2 double bonds. On carbons 9 and 12.

Lets see figure 1

I hope it helps!

5 0
3 years ago
If your car gets 30.3 mi/gal, how many gallons of gasoline would you use if you drove 506.3 miles? 1 mile = 5280 ft (exactly) In
Salsk061 [2.6K]

Answer: 16.7 gallon

Explanation:

Given: The car can drive 30.3 miles when 1 gallon of gasoline is used.

Distance covered = 506.3 miles

Thus for 30.3 miles, the amount of gasoline used= 1 gallon

For 506.3 miles, the amount of gasoline used=\frac{1}{30.3}\times 506.3=16.7gallon

Thus the amount of gasoline used is 16.7 gallons.


4 0
3 years ago
Read 2 more answers
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