Ask question

Ask question

All categories

3 years ago

6

Which kind of URL suffix would you most likely find attached to a site maintained by the Red Cross?

1 answer:

Ostrovityanka [42]3 years ago

7 0

It would most likely be .org or .com :)
Your welcome :D

You might be interested in

Imagine you are a civil engineer who designs buildings. How might testing small-scale building models or their components be hel

lubasha [3.4K]

The error would be 40 cm

3 0

3 years ago

Read 2 more answers

Does the amount of Borax added change the slime?

mixer [17]

Yes. Racist it will make it harder to move almost like ooblek

7 0

3 years ago

Properties of group 7 (aka. The Halogens)?

ella [17]

Halogens can be characterized as :

1) High Electronegativity
2) Low melting & boiling points
3) Forms Ionic compounds
4) Non-metallic in nature

Hope this helps!

7 0

3 years ago

The coefficients in a balanced chemical equation represent

Dafna1 [17]

The coefficients in a chemical equation represent the molar ratio of the substances.

For example, if an equation says 2H2 + O2 ⇒ 2H2O, it means
2 moles of H2 + 1 mol of O2 ⇒ 2 moles of H2O.

8 0

2 years ago

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]

<u>Answer:</u>

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

<u>For B:</u> The $K_c$ for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

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

<u>For A:</u>

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 $4.0\times 10^1$

<u>For B:</u>

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

2 years ago