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3 years ago

Write the complementary DNA strand to A-C-C-G-T-T-A-C-G-C-A

Chemistry

2 answers:

barxatty [35]3 years ago

6 0

Answer:T-G-G-C-A-A-T-G-C-G-T

Explanation:adenine always pairs with thymine and thymine always pairs with adenine

Cytosine always pairs with guanine and guanine always pairs with cytosine

posledela3 years ago

3 0

Answer:

T-G-G-C-A-A-T-G-C-G-T

Explanation:

use AT GC

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If the sentence is true, write true. If the sentence is false , replace the underlined word to make the sentence true.

ivann1987 [24]

1. False

Elements bond to form compounds.

Consider that compounds are essentially clumped up atoms. Knowing this, we know that atoms don’t separate, but rather combine in order to make compounds.

2. True

3. False

Atoms cam lose or gain electrons to form ionic bonds.

When at atom doesn’t have enough electrons to become stable, it will either give or take electrons from another atom in order to become stable. However, because of the fact that the atoms become oppositely charge, they attract each other, thus forming an ionic bond

-T.B.

3 0

3 years ago

How much time would it take for 12000 atoms of nitrogen-13 (half-life is 7.13 s) to decay to 750 atoms?

Rainbow [258]

<h3>Answer:</h3>

28.52 seconds

<h3>Explanation:</h3>

Initial number of atoms of Nitrogen 12,000 atoms

Half-life = 7.13

Number of atoms after decay = 750 atoms

We are required to determine the time taken for the decay.

Note that half life is the time taken for a radioactive isotope to decay to a half of its original amount.

Using the formula;

Remaining amount = Initial amount × (1/2)^n , where n is the number of half lives

In our case;

750 atoms = 12,000 atoms × (1/2)^n

0.0625 = 0.5^n

n = log 0.0625 ÷ log 0.5

n = 4

But, 1 half life =7.13 seconds

Therefore;

Time taken = 7.13 seconds × 4

= 28.52 seconds

Therefore, the time taken for 12,000 atoms of nitrogen to decay to 750 atoms is 28.52 seconds

8 0

3 years ago

Air containing 20.0 mol% water vapor at an initial pressure of 1 atm absolute is cooled in a 1- liter sealed vessel from 200°C t

chubhunter [2.5K]

Answer:

This solution is quite lengthy

Total system = nRT

n was solved to be 0.02575

nH20 = 0.2x0.02575

= 0.00515

Nair = 0.0206

PH20 = 0.19999

Pair = 1-0.19999

= 0.80001

At 15⁰c

Pair = 0.4786atm

I used antoine's equation to get pressure

The pressure = 0.50

2. Moles of water vapor = 0.0007084

Moles of condensed water = 0.0044416

Grams of condensed water = 0.07994

Please refer to attachment. All solution is in there.

6 0

3 years ago

How many moles are present in 356.4 g of NiBr3

lozanna [386]

1.194 mol

(remember to use sig figs!)

8 0

2 years ago

A solution contains [Ba2+] = 5.0 × 10−5 M, [Zn2+] = 2.0 × 10−7 M, and [Ag+] = 3.0 × 10−5 M. Sodium oxalate (Na2C2O4) is slowly a

Jet001 [13]

Answer:

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄

Explanation:

1. Calculate the equilibrium concentrations of oxalate ion

Let [C₂O₄²⁻] = c

(a) Barium oxalate

BaC₂O₄ ⇌ Ba²⁺ + C₂O₄²⁻

E/mol·L⁻¹: 5.0 × 10⁻⁵ c

Ksp = [Ba²⁺][C₂O₄²⁻] = 5.0 × 10⁻⁵c = 1.5 × 10⁻⁸

c = (1.5 × 10⁻⁸)/(5.0 × 10⁻⁵) = 3.0 × 10⁻⁴ mol·L⁻¹

(b) Zinc oxalate

ZnC₂O₄ ⇌ Zn²⁺ + C₂O₄²⁻

E/mol·L⁻¹: 2.0 × 10⁻⁷ c

Ksp = [Zn²⁺][C₂O₄²⁻] = 2.0 × 10⁻⁷c = 1.35 × 10⁻⁹

c = (1.35 × 10⁻⁹)/(2.0 × 10⁻⁷) = 6.8 × 10⁻³ mol·L⁻¹

(c) Silver oxalate

Ag₂C₂O₄ ⇌ 2Ag⁺ + C₂O₄²⁻

E/mol·L⁻¹: 3.0 × 10⁻⁵ c

Ksp = [Ag⁺]²[C₂O₄²⁻] = (3.0× 10⁻⁵)²c = 9.0 × 10⁻¹⁰c = 1.1 × 10⁻¹¹

c = (1.1 × 10⁻¹¹)/(9.0 × 10⁻¹⁰) = 0.012 mol·L⁻¹

2. Decide the order of precipitation

BaC₂O₄ will precipitate when c > 3.0 × 10⁻⁴ mol·L⁻¹

ZnC₂O₄ will precipitate when c > 6.8 × 10⁻³ mol·L⁻¹

Ag₂C₂O₄ will precipitate when c > 0.028 mol·L⁻¹

This happens to be the order of increasing concentration of oxalate ion.

The order of precipitation is

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄

4 0

3 years ago