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beks73 [17]
3 years ago
6

Before Mendeleev published his periodic table, Döbereiner grouped elements with similar properties into "triads," in which the u

nknown properties of one member could be predicted by averaging known values of the properties of the others. To test this idea, predict the values of the following quantities:
(a) The atomic mass of K from the atomic masses of Na and Rb
(b) The melting point of Br₂ from the melting points of Cl₂ (-101.0°C) and I₂ (113.6°C) (actual value = - 7.2°C)
Chemistry
1 answer:
mamaluj [8]3 years ago
4 0

Answer:

a) The atomic mass of the potassium is 54.23 amu.

b) The melting point of bromine gas is 6.3°C.

Explanation:

a) Atomic mass of Na =22.99 amu

Atomic mass of Rb = 85.47 amu

Döbereiner triad = Na , K ,Rb

Taking average of  atomic masses of  Na and Rb

Atomic mass of the K = \frac{22.99 amu+85.47 amu}{2}=54.23 amu

The atomic mass of the potassium is 54.23 amu.

b) Melting point of chlorine gas =-101.0°C

Melting point of iodine gas =113.6°C

Döbereiner triad = Cl, Br , I

Melting point of bromine gas :

=\frac{-101.0^oC +113.6^oC}{2}=6.3^oC

The melting point of chlorine gas is 6.3°C.

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Which two changes would make this reaction reactant-favored?
Vera_Pavlovna [14]

Two changes would make this reaction reactant-favored

C. Increasing the temperature

D. Reducing the pressure

<h3>Further explanation</h3>

Given

Reaction

2H₂ + O₂ ⇒ 2H₂0 + energy

Required

Two changes would make this reaction reactant-favored

Solution

The formation of H₂O is an exothermic reaction (releases heat)

If the system temperature is raised, then the equilibrium reaction will reduce the temperature by shifting the reaction in the direction that requires heat (endotherms). Conversely, if the temperature is lowered, then the equilibrium shifts to a reaction that releases heat (exothermic)  

While on the change in pressure, then the addition of pressure, the reaction will shift towards a smaller reaction coefficient  

in the above reaction: the number of coefficients on the left is 3 (2 + 1) while the right is 2

As the temperature rises, the equilibrium will shift towards the endothermic reaction, so the reaction shifts to the left towards H₂ + O₂( reactant-favored)

And reducing the pressure, then the reaction shifts to the left H₂ + O₂( reactant-favored)⇒the number of coefficients is greater

4 0
2 years ago
Determine the percentage composition (by mass) of oxygen in NH4NO3.
Brrunno [24]

Answer:

60%

Explanation:

M(NH4NO3) = 2*14 +4*1 +3*16 = 80 g/mol

M(3O) = 3*16 = 48 g/mol

(48/80) *100 % =60% oxygen by mass.

6 0
2 years ago
20 POINTS! FAST 1. Write a generic word equation and a chemical equation for the reaction that occurs
Anna35 [415]
I don’t know about 1 But number 2 is AB + xH2O = AB.xH2O
5 0
3 years ago
0.0027432 using scientific notation
alexira [117]

Answer: 2.7423 × 10^-3

Explanation:

4 0
3 years ago
Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the abo
Romashka-Z-Leto [24]

Answer:

\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}

Explanation:

Volume of a cone:

  • \displaystyle V=\frac{1}{3} \pi r^2 h

We have \displaystyle \frac{dV}{dt} = \frac{10 \ cm^3}{sec} and we want to find \displaystyle \frac{dh}{dt} \Biggr | _{h\ =\ 6}= \ ? when the height is 2 cm.

We can see in our equation for the volume of a cone that we have three variables: V, r, and h.

Since we only have dV/dt and dh/dt, we can rewrite the equation in terms of h only.

We are given that the height of the cone is 1/5 the radius at any given time, 1/5r, so we can write this as r = 5h.

Plug this value for r into the volume formula:

  • \displaystyle V =\frac{1}{3} \pi (5h)^2 h  
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Differentiate this equation with respect to time t.

  • \displaystyle \frac{dV}{dt}  =\frac{25}{3} \pi \ 3h^2 \ \frac{dh}{dt}
  • \displaystyle \frac{dV}{dt}  =25 \pi h^2 \ \frac{dh}{dt}

Plug known values into the equation and solve for dh/dt.

  • \displaystyle 10 = 25 \pi (2)^2  \ \frac{dh}{dt}
  • \displaystyle 10 = 100 \pi  \ \frac{dh}{dt}  

Divide both sides by 100π to solve for dh/dt.

  • \displaystyle \frac{10}{100 \pi} = \frac{dh}{dt}
  • \displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}

The height of the cone is increasing at a rate of 1/10π cm per second.

7 0
2 years ago
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