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-Dominant- [34]
3 years ago
11

Ribosomes are cell structures that make

Chemistry
2 answers:
Vitek1552 [10]3 years ago
8 0

Answer:

Ribosomes preform biological synthesis. (mRNA Translation). They also link amino acids together to form polypeptide chain.

kvv77 [185]3 years ago
4 0
Ribosomes link amino acids together in the order specified by the codons of messenger RNA (mRNA) molecules to form polypeptide chains.
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I need to know the following above
arsen [322]
3Zn + 8HNO3 ---> 3Zn(NO3)2 + 4H2O + 2NO IF IT IS COLD AND DILUT NITRIC ACID .

IF IT IS HOT AND CONCENTRATED THEN:

Zn+ 4HNO3 ---> Zn(NO3)2 +2H2O +2NO2
6 0
3 years ago
What does the "R-" represent?
lozanna [386]

Answer:

<em><u>general formula RCOX, where R represents an alkyl or aryl organic radical group, CO ... represents a halogen atom such as chlorine ... loss of a hydroxyl group (-OH), viz, acetyl,. CH, CO- ..</u></em>

8 0
3 years ago
.What threatens the newly hatched chicks?
marysya [2.9K]

Answer:

rats. that's all i know of Just about everything except the mother hen if they are natural hatch. Even when you incubate them there are threats. The healthy chicks will mob the weak ones, the older chicks (even by a day) will pick on the younger ones. Temperature extremes will threaten them as they need warm, humid conditions with gradual drops in surrounding temps in the brooder box. Early disease is sometimes a problem and all chicks should be started on medicated chick feed for the first few weeks to prevent several digestive diseases. Even the water dispenser can be a threat as newly hatched chicks will immerse themselves in an open water container so care should be taken to supply water in a self feeding covered dish.

Explanation:

3 0
3 years ago
Read 2 more answers
Ammonia NH3 may react with oxygen to form nitrogen gas and water.4NH3 (aq) + 3O2 (g) \rightarrow 2 N2 (g) + 6H2O (l)If 2.15g of
bagirrra123 [75]

Answer:

NH3 is the limiting reactant

The % yield is 36.1 %

Explanation:

<u>Step 1: </u>Data given

Mass of NH3 = 2.15 grams

Mass of O2 = 3.23 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

volume of N2 produced = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

<u>Step 2:</u> The balanced equation:

4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)

<u>Step 3:</u> Calculate moles of NH3

Moles NH3 = Mass NH3 / Molar Mass NH3

Moles NH3 = 2.15 grams / 17.03 g/mol

Moles NH3 = 0.126 moles

<u>Step 4:</u> Calculate moles of O2

Moles O2 = 3.23 grams / 32 g/mol

Moles O2 = 0.101 moles

<u>Step 5: </u>Calculate the limiting reactant

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

There will remain 0.101 - 0.945 = 0.0065 moles of O2

<u>Step 6:</u> Calculate moles of N2

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

<u>Step 7</u>: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

<u>Step 8:</u> Calculate the % yield

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

4 0
3 years ago
describe how a physical property such as mass or Texture can change without causing a change in the substance
Tom [10]

You have to be very careful with this question. A change in mass can also occur in chemical changes especially if you have too much of something. For example

CH4 + 1.5 02 ===> CO2 + H2O

If you have too much of either CH4 or O2, there will be some CH4 or O2 left over. There has been a change in mass that you have too much of.

However that is not the point of the question. It is just something you need to be aware of.

Suppose you have a piece of aluminum and you take a course grinder after it. You will change the texture of the side you took the grinder to. If the aluminum has been anodized (a color has been put on it's surface), you may grind the color off or if it is just plain aluminum, you may roughen the surface, but you won't change what the aluminum will do chemically.

You may need only a small portion of the aluminum and you grind off just what you need. That will change the mass of both what you took off and the piece that you want, but the aluminum will still do whatever chemical property you need to use.

So you can change both texture and mass without changing the chemical properties of the substance whose mass or texture you are changing.

3 0
3 years ago
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