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3 years ago

How many copies of DNA would be created after 24 cycles?

Chemistry

1 answer:

vaieri [72.5K]3 years ago

5 0

Answer:

After 30 cycles, as many as a billion copies of the target sequence are produced from a single starting molecule

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The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

3 years ago

Can someone please help me

Lilit [14]

It is A. Barium

Explanation: I did that already

3 0

3 years ago

A substance that loses electron and itself is oxidized is called a?

Verdich [7]

Answer:

oxidation-reduction or redox reaction

Explanation:

8 0

3 years ago

Read 2 more answers

Determine the resulting pH when 12mL if 0.16M HCl are reacted with 32 mL if 0.24M KOH.

TEA [102]

Answer:

pH = 13.1

Explanation:

Hello there!

In this case, according to the given information, we can set up the following equation:

$HCl+KOH\rightarrow KCl+H_2O$

Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:

$n_{HCl}=0.012L*0.16mol/L=0.00192mol\\\\n_{KOH}=0.032L*0.24mol/L=0.00768mol$

Thus, since there are less moles of HCl, we calculate the remaining moles of KOH as follows:

$n_{KOH}=0.00768mol-0.00192mol=0.00576mol$

And the resulting concentration of KOH and OH ions as this is a strong base:

$[KOH]=[OH^-]=\frac{0.00576mol}{0.012L+0.032L}=0.131M$

And the resulting pH is:

$pH=14+log(0.131)\\\\pH=13.1$

Regards!

3 0

3 years ago

A 1.00 liter solution contains 0.24 M hypochlorous acid and 0.31 M sodium hypochlorite. If 0.160 moles of potassium hydroxide ar

ryzh [129]

Explanation:

When OH- (as in potassium hydroxide) is added, it reacts with the acid (HOCl) to reduce the amount of HOCl and increase the concentration of sodium hypochlorite.

Potassium hydroxide will react with the hypochlorous acid to produce hypochlorite ions. In the process, some of the weak acid will be consumed, along with the added strong base.

This occurs as follows:

HClO(aq) + KOH(aq) → KClO(aq) + H2O(l)

since water is formed, this maintains the pH. Thus ...

A. The number of moles of HClO will decrease. - TRUE

B. The number of moles of ClO- will increase. - TRUE

C. The equilibrium concentration of H3O+ will remain the same. - TRUE

D. The pH will decrease. - FALSE

E. The ratio of [HClO] / [ClO-] will decrease. -TRUE. It will decrease as HClO goes down and ClO- goes up.

5 0

4 years ago