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Stolb23 [73]
3 years ago
13

_al + _h2o -> al(oh)3 + _h2

Chemistry
1 answer:
Mazyrski [523]3 years ago
7 0

The answer is 2 6 2 6

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On a very hot day, could you cool your kitchen by leaving the refrigerator door open?
Scilla [17]

Answer:c

Explanation:

8 0
3 years ago
Read 2 more answers
How much heat must be removed from 456 g of water at 25.0°c to change it into ice at?
N76 [4]
This question can be simply solved by using heat formula,
    Q = mCΔT

Q = heat energy (J)
m = Mass (kg)
C = Specific heat capacity (J / kg K)
ΔT = Temperature change (K)

when water freezes, it produces ice at 0°C (273 K)
hence the temperature change is 25 K (298 K - 273 K)
C for water is 4186 J / kg K or 4.186 J / g K
By applying the equation,
 Q = 456 g x 4.186 J / g K x 25 K
     = 47720.4 J
     = 47.72 kJ

hence 47.72 kJ of heat energy should be removed.
5 0
3 years ago
Which statements describe Earth's continents? Check all that apply.
kari74 [83]

Answer:

A continent is a large distinct landmass.

There are seven continents on Earth.

Continents cover two-thirds of Earth’s surface.

Explanation:

8 0
3 years ago
The half-life of tritium (H-3) is 12.3 years. If 48.0mg of tritium is released from a nuclear power plant during the course of a
Rudiy27

Answer:

The amount left after 49.2 years is 3mg.

Explanation:

Given data:

Half life of tritium = 12.3 years

Total mass pf tritium = 48.0 mg

Mass remain after 49.2 years = ?

Solution:

First of all we will calculate the number of half lives.

Number of half lives = T elapsed/ half life

Number of half lives =  49.2 years /12.3 years

Number of half lives =  4

Now we will calculate the amount left after 49.2 years.

At time zero 48.0 mg

At first half life = 48.0mg/2 = 24 mg

At second half life = 24mg/2 = 12 mg

At 3rd half life = 12 mg/2 = 6 mg

At 4th half life =  6mg/2 = 3mg

The amount left after 49.2 years is 3mg.

6 0
3 years ago
Consider the reaction: P(s) + 5/2 Cl2(g)PCl5(g) Write the equilibrium constant for this reaction in terms of the equilibrium con
Pani-rosa [81]

Answer: The equilibrium constant for the overall reaction is K_a\times K_b

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.

a) P(s)+\frac{3}{2}Cl_2(g)\rightarrow PCl_3(g)

K_a=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}

b) PCl_3(g)+Cl_2(g)\rightarrow PCl_5(g)

K_b=\frac{[PCl_5]}{[Cl_2]\times [PCl_3]}

For overall reaction on adding a and b we get c

c) P(s)+\frac{5}{2}Cl_2(g)\rightarrow PCl_5(g)

K_c=\frac{[PCl_5]}{[Cl_2]^\frac{5}{2}}

K_c=K_a\times K_b=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}\times \frac{[PCl_5]}{[Cl_2]\times [PCl_3]}

The equilibrium constant for the overall reaction is K_a\times K_b

4 0
3 years ago
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