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2 years ago

Dna is made up of nitrogenous base pairs. the bases bond in which of the following patterns?

1 answer:

6 0

The nitrogenous bases on the two strands of DNA pair up, purine with pyrimidine (A with T, G with C), and are held together by weak hydrogen bonds. Watson and Crick discovered that DNA had two sides, or strands, and that these strands were twisted together like a twisted ladder -- the double helix.Mar

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8) Determine whether mixing each pair of the following results in a buffera. 100.0 mL of 0.10 M NH3 with 100.0 mL of 0.15 MNH4Cl

Marina86 [1]

Answer:

a. 100.0 mL of 0.10 M NH₃ with 100.0 mL of 0.15 M NH₄Cl.

c. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOH.

Explanation:

A buffer system is formed in 1 of 2 ways:

A weak acid and its conjugate base.

A weak base and its conjugate acid.

Determine whether mixing each pair of the following results in a buffer.

a. 100.0 mL of 0.10 M NH₃ with 100.0 mL of 0.15 M NH₄Cl.

YES. NH₃ is a weak base and NH₄⁺ (from NH₄Cl ) is its conjugate base.

b. 50.0 mL of 0.10 M HCl with 35.0 mL of 0.150 M NaOH.

NO. HCl is a strong acid and NaOH is a strong base.

c. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOH.

YES. HF is a weak acid and it reacts with NaOH to form NaF, which contains F⁻ (its conjugate base).

d. 175.0 mL of 0.10 M NH₃ with 150.0 mL of 0.12 M NaOH.

NO. Both are bases.

6 0

2 years ago

The vapor above a mixture of pentane and hexane at room temperature contains 35.5% pentane by mass. What is the mass percent com

FrozenT [24]

The mass percent of Pentane in solution is 16.49%

The mass percent of Hexane in solution is 83.51%

<u>Explanation</u>:

Take 1 kg basis for the vapor: 35.5 mass% pentane = 355 g pentane with 645 g hexane.

Convert these values to mol% using their molecular weights:

Pentane: Mp = 72.15 g/mol -> 355g/72.15 g/mol = 4.92mol

Hexane: Mh = 86.18 g/mol -> 645g/86.18 g/mol = 7.48mol

Pentane mol%: yp = 4.92/(4.92+7.48) = 39.68%

Hexane mol%: yh = 100 - 39.68 = 60.32%

Pp-vap = 425 torr = 0.555atm

Ph-vap = 151 torr = 0.199atm

From Raoult's law we know:

Pp = xp $\times$ Pp - vap = yp $\times$ Pt (1)

Ph = xh $\times$ Ph - vap = yh $\times$ Pt (2)

Since it is a binary mixture we can write xh = (1 - xp) and yh = (1 - yp), therefore (2) becomes:

(1 - xp) $\times$ Ph - vap = (1 - yp) $\times$ Pt (3)

Substituting (1) into (3) we get:

(1-xp) $\times$ Ph - vap = (1 - yp) $\times$ xp $\times$ Pp - vap / yp (4)

Rearrange for xp:

xp = Ph - vap / (Pp - vap/yp - Pp - vap + Ph - vap) (5)

Subbing in the values we find:

Pentane mol% in solution: xp = 19.08%

Hexane mol% in solution: xh = 80.92%

Now for converting these mol% to mass%, take 1 mol basis for the solution and multiplying it by molar mass:

mp = 0.1908 mol $\times$ 72.15 g/mol

= 13.766 g

mh = 0.8092 mol $\times$ 86.18 g/mol

= 69.737 g

Mass% of Pentane solution = 13.766/(13.766+69.737)

= 16.49%

Mass% of Hexane solution = 83.51%

8 0

3 years ago

Plz help me I am!!!!

Soloha48 [4]

Gravity forces earth to move around the sun so A

8 0

3 years ago

Read 2 more answers

Consider the reaction 2 NO + O2 → 2 NO2

Sonbull [250]

Answer:

(a) Rate at which $NO_2$ is formed is 0.050 M/s

(b) Rate at which $O_2$ is consumed is 0.0250 M/s.

Explanation:

The given reaction is:-

$2NO+O_2\rightarrow 2NO_2$

The expression for rate can be written as:-

$-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}$

Given that:- $\frac{d[NO]}{dt}=-0.050\ M/s$ (Negative sign shows consumption)

$-\frac{1}{2}\frac{d[NO]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}$

$-\frac{d[NO]}{dt}=\frac{d[NO_2]}{dt}$

$-(-0.050\ M/s)=\frac{d[NO_2]}{dt}$

$\frac{d[NO_2]}{dt}=0.050\ M/s$

(a) Rate at which $NO_2$ is formed is 0.050 M/s

$-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}$

$-\frac{1}{2}\times -0.050\ M/s=-\frac{d[O_2]}{dt}$

$\frac{d[O_2]}{dt}=0.0250\ M/s$

(b) Rate at which $O_2$ is consumed is 0.0250 M/s.

5 0

3 years ago

A 100.0 g sample of aluminum released 1680 calories when cooled from 100.0c to 20.0 c what is the specific heat of the metal

vovangra [49]

Answer:

The specific heat of aluminium is 0.8792 J/g °C or 0.21 Cal/g °C

Explanation:

Step 1 : Write formule of specific heat

Q=mcΔT

with Q = heat transfer (J)

with m = mass of the substance

with c = specific heat ⇒ depends on material and phase ( J/g °C)

with ΔT = Change in temperature

For this case :

Q = 1680 Calories = 7033.824 J ( 1 calorie = 4.1868 J)

m = 100.0g

c= has to be determined

ΔT = 100 - 20 = 80°C

<u>Step 2: Calculating specific heat</u>

⇒ via the formule Q=mcΔT

7033.824 J = 100g * c * 80

7033.824 = 8000 *c

c = 7033.824 /8000

c = 0,879228 J/g °C

or 0.21 Cal / g°C

The specific heat of aluminium is 0.8792 J/g °C or 0.21 Cal/g °C

6 0

3 years ago