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DiKsa [7]
3 years ago
11

Where is the genetic code located again in the dna molecule again?

Chemistry
1 answer:
Pie3 years ago
8 0

Answer:

The DNA is in the nucleus.

Explanation:

The nucleus is the power house of the cell, the little sting things inside are the strands of DNA.

You might be interested in
In the following reaction, 451.4 g of lead reacts with excess oxygen forming 365.0 g of lead(II) oxide. Calculate the percent yi
user100 [1]
Let MM(x) be the molar mass of x.

MM(Pb) : MM(PbO)
=207.21 : 223.20  =  451.4 g : x g

cross multiply and solve for x
x=223.2/207.21*451.4
= 486.23 g

Percentage yield = 365.0/486.23= 0.75067 = 75.07% (rounded to 4 sign. fig.)

4 0
3 years ago
Mg(OH)2+2 HCI = MgCI2+ H2O
In-s [12.5K]

Answer:

Mg(OH)2 + 2HCI ⇒ MgCI2 + 2H2O

Explanation:

In order to balance a chemical equation you need to make sure both sides of the equations are equal.

Mg(OH)2 + 2HCI = MgCI2 + H2O

Mg = 1

Oh = 2

HCI = 2

Products:

Mg = 1

CI = 2

H = 2

O = 1

2H20 = 1 × 2 = 2

2 × 2 = 4

2HCI

1 × 2 = 2

Mg(OH)2 + 2HCI ⇒ MgCI2 + 2H2O

Hope this helps.

7 0
3 years ago
When dipentyl ether is treated with HI, what type of reaction occurs? both SN1 and SN2 SN2 E1 SN1 E2
masha68 [24]

Answer:

SN2

Explanation:

The first step of ether cleavage is the protonation of the ether since ROH is a better leaving group than RO-.

The second step of the reaction may proceed by either SN1 or SN2 mechanism depending on the structure of the ether. Methyl and primary ethers react with HI by SN2 mechanism while tertiary ethers react with HI by SN1 mechanism. Secondary ethers react with HI by a mixture of both mechanisms.

Dipentyl ether is a primary ether hence when treated with HI, the reaction with HI proceeds by SN2 mechanism as explained above.

7 0
3 years ago
How many grams in 10 ounces
Kamila [148]

Answer:

283.495g

Explanation:

8 0
3 years ago
Read 2 more answers
At 25°c the henry's law constant for nitrogen trifluoride (nf3) gas in water is 7.9 × 10-4 m/atm. what is the mass of nf3 gas th
Step2247 [10]
For this problem, we should use the Henry's Law formula which is written below:

P = kC
where
P is the partial pressure of the gas
k is the Henry's Law constant at a certain temperature
C is the concentration

Substituting the values,
1.71 atm = (7.9×10⁻⁴<span> /atm)C
Solving for C,
C = 2164.56 molal or 2164.56 mol/kgwater

Let's make use of density of water (</span>1 kg/1 m³) and the molar mass of NF₃ (71 g/mol).<span>

Mass of NF</span>₃ = 2164.56 mol/kg water * 1 kg/1 m³ * 1 m³/1000000 mL * 150 mL * 71 g/mol = 23.05 g 
4 0
3 years ago
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