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user100 [1]
3 years ago
12

If a student rode his bike 102 hectometers, how many kilometers did he ride?

Chemistry
1 answer:
Maru [420]3 years ago
4 0
10.2 kilometers is how long he would ride
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Paano sabihin kay crush na crush mo siya?​
aleksley [76]

Answer:

mean

Explanation:

acording to whom chrush

4 0
2 years ago
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) determine the henry's law constant for ammonia in water at 25°c if an ammonia pressure of 0.022 atm produces a solution with a
Nataly [62]

Answer:

a. 59 m/atm

Explanation:

  • To solve this problem, we must mention Henry's law.
  • <em>Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.</em>
  • It can be expressed as: C = KP,

C is the concentration of the solution (C = 1.3 M).

P is the partial pressure of the gas above the solution (P = 0.022 atm).

K is the Henry's law constant (K = ??? M/atm),

∵ C = KP.

∴ K = C/P = (1.3 M)/(0.022 atm) = 59.0 M/atm.

3 0
3 years ago
What stress would shift the equilibrium position of the following system to the left?
Margarita [4]

Answer:

Heating the system

Explanation:

According to the principle of Le Chatelier, for a system at equilibrium, a specific disturbance would make the equilibrium shift toward the direction which minimizes such a disturbance.

Since we wish to shift the equilibrium to the left, this means we wish to increase the concentration of products, as an excess in their concentration would make the products react and produce more reactants in order to lower the excess concentration of products.

Since heat is also a product, an increase in heat would shift the equilibrium toward the left, as this would consume the excess of heat by producing the reactants.

4 0
3 years ago
(Thermodynamics)
frutty [35]

Answer:

3853 g

Step-by-step explanation:

M_r: 107.87

         16Ag + S₈ ⟶ 8Ag₂S; ΔH°f =  -31.8 kJ·mol⁻¹

1. Calculate the moles of Ag₂S

Moles of Ag₂S = 567.9 kJ × 1 mol Ag₂S/31.8kJ = 17.858 mol Ag₂S

2. Calculate the moles of Ag

Moles of Ag = 17.86 mol Ag₂S × (16 mol Ag/8 mol Ag₂S) = 35.717 mol Ag

3. Calculate the mass of Ag

Mass of g = 35.717 mol Ag × (107.87 g Ag/1 mol Ag) = 3853 g Ag

You must react 3853 g of Ag to produce 567.9 kJ of heat

3 0
3 years ago
Based on the sign of E cell, classify these reactions as spontaneous or non spontaneous as written.? assume standard conditions.
sammy [17]
A electrochemical reaction is said to be spontaneous, if E^{0} cell is positive. 

Answer 1:
Consider reaction: <span>Ni^2+ (aq) + S^2- (aq) ----> + Ni (s) + S (s) 

The cell representation of above reaction is given by;
    </span>S^{2-}/S //  Ni^{2+}/Ni

Hence, E^{0}cell =  E^{0} Ni^{2+/Ni} -  E^{0} S/S^{2-}
we know that, {E^{0} Ni^{2+}/Ni  = -0.25 v
and {E^{0} S/ S^{2-}  = -0.47 v

Therefore, E^{0} cell = - 0.25 - (-0.47) = 0.22 v

Since,  E^{0} cell is positive, hence cell reaction is spontaneous
.....................................................................................................................

Answer 2: 
Consider reaction: <span>Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq)
</span>
The cell representation of above reaction is given by;
    H_{2} /  H^{+} //  Pb^{2+} /Pb

Hence, E^{0}cell = E^{0} Pb/Pb^{2+} - E^{0} H_{2}/H^{+}
we know that, {E^{0} Pb^{2+}/Pb = -0.126 v
and {E^{0} H_{2}/ H^{+} = -0 v

Therefore, E^{0} cell = - 0.126 - 0 = -0.126 v

Since,  E^{0} cell is negative, hence cell reaction is non-spontaneous.

....................................................................................................................

Answer 3: 
Consider reaction: <span>2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq)
</span>
The cell representation of above reaction is given by;
    Cr/Cr^{2+} // Ag^{+}/Ag

Hence, E^{0}cell = E^{0} Ag^{+}/Ag - E^{0} Cr/Cr^{2+}
we know that, {E^{0} Ag^{+}/Ag = -0.22 v
and {E^{0} Cr/ Cr^{2+} = -0.913 v

Therefore, E^{0} cell = - 0.22 - (-0.913) = 0.693 v

Since,  E^{0} cell is positive, hence cell reaction is spontaneous
8 0
3 years ago
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