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timurjin [86]
3 years ago
11

If I was traveling 76 mph on the highway, how many ly/s am I going?

Chemistry
1 answer:
Sphinxa [80]3 years ago
5 0

Answer:

the speed in light-year per second is 2.155 x 10⁻¹³ ly/s.

Explanation:

Given;

speed in miles per hour (mph), v = 76 mph

1 mile = 1.701 x 10⁻¹³ light-year (ly)

1 hour = 60 s

The speed in light-year per second is calculated as follows;

v = \frac{76 \ miles}{hour} \times \frac{1.701 \times 10^{-13} \ ly}{1 \ mile} \times \frac{1 \ hour}{60 \ s} = 2.155 \times 10^{-13} \ ly/s

Therefore, the speed in light-year per second is 2.155 x 10⁻¹³ ly/s.

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C. Oxide is a simple radical why??<br>​
Luda [366]

Answer:

In chemistry, a radical is an atom, molecule, or ion that has an unpaired valence electron. ... In living organisms, the radicals super oxide and nitric oxide and their reaction products regulate many processes

Explanation:

7 0
3 years ago
Write the structure of all possible peptides containing these amino acids: Asp, Glu, Gln Use single letter abbreviations and cap
Liula [17]

Answer:

QED,  EDQ,  DQE,  EQD,  DEQ,  QDE

Explanation:

The structure of all possible peptides that contain the given amino acids are :

QED,  EDQ,  DQE,  EQD,  DEQ,  QDE

where : Asp is represented by the letter code D

             Glu is represented by the letter code E

             Gln is represented by the letter code Q

Note : when three amino acids combine they form what is known as tripeptide ( i.e. contains two peptide linkages ) while a peptide linkage is been formed by the combination of a carboxyl group of an amino acid and the amino group of different amino acid

6 0
3 years ago
Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Supposed 0.481 g of methane
Shalnov [3]

Answer:

1.08g

Explanation:  

Like all other hydrocarbons, methane burns in oxygen to form carbon iv oxide and water. The chemical equation of this equation is shown below;

  CH4 (g)  +  2O2  (g) ----------->  2H2O (l)   +   CO2 (g)

From the reaction equation, we can see that one mole of methane gave 2  moles of water. This is the theoretical yield. We need to note the actual yield.

To get the actual yield, we get the number of moles of methane reacted. To get this, we divide the mass of methane by  the molar mass of methane. The molar mass of methane is 16g/mol. The number of moles is thus 0.481/16 = 0.03 moles.

Since 1 mole methane gave 2 moles of water, this shows that 0.06 moles of water were produced. The mass of water thus produced is 0.06 multiplied by the molar mass of water. The molar mass of water is 18g/mol. The mass produced is 0.06 * 18 = 1.08g

Now, we do same for the mass of oxygen. From the reaction equation, 2 moles of oxygen produced two moles of water. Hence, we can see from here that the number of moles here are equal. We then proceed to calculate the actual number of moles of oxygen produced. This is the mass of the oxygen divided by the molar mass of molecular oxygen. The molar mass of molecular oxygen is 32g/mol. The number of moles thus produced is 0.54/32 = 0.016875 mole. The number of moles are equal, this means that the number of moles of oxygen produced is also 0.016875

Now, to get the mass of water produced, we multiply the number of moles by the molar mass of water. The molar mass of water is 18g/mol.

The mass is thus, 0.016875 * 18 = 0.30375g

1.08g is higher and thus is the maximum mass

3 0
3 years ago
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Olin [163]
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A metal weighing 7.101 g is placed in a graduated cylinder containing 33.0 mL of water. The water level rose to the 37.4 mL mark
musickatia [10]
Volume of the metal = change in volume reading
Volume = 37.4 - 33
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Density = mass / volume
Density = 7.101 / 4.4
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7 0
3 years ago
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