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Gold is formed in pure state in nature but not iron,why?

ss7ja [257]

Gold is found in pure state in nature but not the iron because gold is un-reactive metal, so it does not react with other element in normal condition but iron is reactive metal so it reacts with atmospheric oxygen in presence of moist (water) to form oxide.

Explanation:

I hope it'll help you.

5 0

2 years ago

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Calculate the solubility of o2 in water at a partial pressure of o2 of 120 torr at 25 ̊c. the henry's law constant for o2 at 25

Vladimir79 [104]

Answer:

1) 2.054 x 10⁻⁴ mol/L.

2) Decreasing the temperature will increase the solubilty of O₂ gas in water.

Explanation:

1) The solubility of O₂ gas in water:

We cam calculate the solubility of O₂ in water using Henry's law: <em>Cgas = K P</em>,

where, Cgas is the solubility if gas,

K is henry's law constant (K for O₂ at 25 ̊C is 1.3 x 10⁻³ mol/l atm),

P is the partial pressure of O₂ (P = 120 torr / 760 = 0.158 atm).

Cgas = K P = (1.3 x 10⁻³ mol/l atm) (0.158 atm) = 2.054 x 10⁻⁴ mol/L.

2) The effect of decreasing temperature on the solubility O₂ gas in water:

Decreasing the temperature will increase the solubilty of O₂ gas in water.

When the temperature increases, the solubility of O₂ gas in water will decrease because the increase in T will increase the kinetic energy of gas particles and increase its motion that will break intermolecular bonds and escape from solution.

Decreasing the temperature will increase the solubility of O₂ gas in water will because the kinetic energy of gas particles will decrease and limit its motion that can not break the intermolecular bonds and increase the solubility of O₂ gas.

6 0

2 years ago

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The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that

kakasveta [241]

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is $1.1\times 10^6$ M.

$HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)$

This value indicates that

A. $CN^-$ is a stronger base than $ONO^-$

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is $ONO^-$

D. The conjugate acid of CN- is HCN

Answer: A. $CN^-$ is a stronger base than $ONO^-$

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When $K_{p}>1$ ; the reaction is product favoured.

When $K_{p};$ ; the reaction is reactant favored.

$When K_{p}=1$ ; the reaction is in equilibrium.

As, $K_p>>1$ , the reaction will be product favoured and as it is a acid base reaction where $HONO$ acts as acid by donating $H^+$ ions and $CN^-$ acts as base by accepting $H^+$

Thus $HONO$ is a strong acid thus $ONO^-$ will be a weak conjugate base and $CN^-$ is a strong base which has weak $HCN$ conjugate acid.

Thus the high value of K indicates that $CN^-$ is a stronger base than $ONO^-$

7 0

3 years ago

Consider three gases: Ar, SF6, and Cl2. If 50.0 grams of these gases are placed in each of three identical containers, which con

adoni [48]

The ideal gas law:
$pV=nRT \Rightarrow p=\frac{nRT}{V}$
p - pressure, n - number of moles, R - the gas constant, T - temperature, V - volume

The volume and temperature of all three containers are the same, so the pressure depends on the number of moles. The greater the number of moles, the higher the pressure.
The mass of gases is 50 g.

$Ar \\
M \approx 39.948 \ \frac{g}{mol} \\
n=\frac{50 \ g}{39.948 \ \frac{g}{mol}} \approx 1.25 \ mol \\ \\
SF_6 \\
M \approx 146.06 \ \frac{g}{mol} \\
n=\frac{50 \ g}{146.06 \ \frac{g}{mol}} \approx 0.34 \ mol \\ \\
Cl_2 \\
M=70.9 \ \frac{g}{mol} \\
n=\frac{50 \ g}{70.9 \ \frac{g}{mol}} \approx 0.71 \ mol$

The greatest number of moles is in the container with Ar, so there is the highest pressure.

4 0

3 years ago

Help! 15 Points!<br><br> Science!

Klio2033 [76]

Don't take my word for it but I think it is
1: proteins
2: energy from the sun, carbon dioxide, and water
3: this one is confusing me but I think it would be nutrients from food and oxygen
4: water
Again these are attempts I can't prove these

6 0

3 years ago

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