Which event always involves a chemical change A boiling B melting C conducting D burning

Brilliant_brown [7]

Answer:

The answer is D if it's wrong let me know pls

5 0

2 years ago

As an employee of a summer sports camp, one of your jobs is to prepare the tank of vitamin water for use throughout the day. the

TiliK225 [7]

Given:
Mixture of vitamin water = 75% pure water & 25% concentrated vitamin drink.
To find:
Quantity of water to be added to 16 gallons of concentrated vitamin drink to prepare a tank of vitamin water.
Solution:
Let total amount of mixture be x.
25% of x = 16 gallons (from given information)
When 25% of x is 16 gallons, the remaining 75% of x will be calculated as below:
(16/25)*75 = 48
Answer: 75% of x = 48 gallons. This means 48 gallons pure water is required to be added to mixture to prepare a tank of vitamin water.

3 0

3 years ago

Open the Balancing Chemical Equations interactive and select Introduction mode. Then choose Separate Water. Adjust the coefficie

aleksklad [387]

Explanation:

1. Water decomposition

Decomposition reactions are represented by-

The general equation: AB → A + B.

Various methods used in the decomposition of water are -

Electrolysis
Photoelectrochemical water splitting
Thermal decomposition of water
Photocatalytic water splitting

Water decomposition is the chemical reaction in which water is broken down giving oxygen and hydrogen.

The chemical equation will be -

$H_2O + H_2 + O_2$

Hence, balancing the equation we need to add a coefficient of 2 in front of $H_2O$ on right-hand-side of the equation and 2 in front of $H_2$ on left-hand-side of the equation.

∴The balanced equation is -

$2 H_2O$ → $2 H_2 + O_2$

2. Formation of ammonia

The formation of ammonia is by reacting nitrogen gas and hydrogen gas.

$N_2 + H$ → $NH_3$

Hence, for balancing equation we need to add a coefficient of 3 in front of hydrogen and 2 in front of ammonia.

∴The balanced chemical equation for the formation of ammonia gas is as follows -

$N_2+3H$ → $2NH_3$ .

When 6 moles of $N_2$ react with 6 moles of $H_2$ 4 moles of ammonia are produced.

5 0

2 years ago

Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1

Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

Explanation:

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of $SbCl_5$ is increasing .So, the equilibrium will shift in the right direction.