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2 years ago

Which is a single ringed nitrogeo base? A)adenine B)guanine C)thymine D)ribose

Chemistry

1 answer:

pogonyaev2 years ago

3 0

A single ringed nitrogeo base is thymine (C)

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The acid dissociation constant Ka of boric acid (H3BO3) is 5.8 times 10^-10. Calculate the pH of a 4.4 M solution of boric acid.

madam [21]

Answer: The pH of a 4.4 M solution of boric acid is 4.3

Explanation:

$H_3BO_3\rightarrow H^+H_2BO_3^-$

at t=0 cM 0 0

at eqm $c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 4.4 M and $\alpha$ = ?

$K_a=5.8\times 10^{-10}$

Putting in the values we get:

$5.8\times 10^{-10}=\frac{(4.4\times \alpha)^2}{(4.4-4.4\times \alpha)}$

$(\alpha)=0.000011$

$[H^+]=c\times \alpha$

$[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M$

Also $pH=-log[H^+]$

$pH=-log[4.8\times 10^{-5}]=4.3$

Thus pH of a 4.4 M $H_3BO_3$ solution is 4.3

3 0

2 years ago

Liquid nitrogen is used as a commercial refrigerant to flash freeze foods. Nitrogen boils at

Arisa [49]

77.15 kelvin would be the answer

5 0

2 years ago

Read 2 more answers

How many C-13 atoms are present, on average, in a 1.6000×104-atom sample of carbon?

san4es73 [151]

I don't know why I am answering this question but assuming C-13 has a natural abundance of 1.07%:

(1.6000x10^4)(0.0107) = 171.2 = 171 atoms of C-13

5 0

2 years ago

Read 2 more answers

Find the volume of 20g of benzene

sergey [27]

(d) Density of Benzene: 0.8786 g/cm cubed

(m) Mass: 20.00g

Formula to solve (v) volume of benzene: V=m/d

V=20g / 0.8786g/cm cubed

Answer: Volume of benzene is: 22.8 cm cubed

Explanation: Well, density is mass divided volume. In this, it is what the volume is. All you need is mass divided by density. Hope this helps!

6 0

2 years ago

Zn-64 = 48.63%

sladkih [1.3K]

Answer:

A = 65.46 u

Explanation:

Given that,

The composition of zinc is as follows :

Zn-64 = 48.63%

Zn-66 = 27.90%

Zn-67 = 4.10%

Zn-68 = 18.75%

Zn-70 = .62%

We need to find the average atomic mass of the given element. It can be solved as follows :

$A=\dfrac{48.63\times 64+27.90\times 66+4.1\times 67+18.75\times 68+0.62\times 70}{100}\\A=65.46\ u$

So, the average atomic mass of zinc is 65.46 u.

6 0

2 years ago