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2 years ago

A mixture consists of sand and an aqueous salt

1 answer:

4 0

The right answer for the question that is being asked and shown above is that: "(1) Evaporate the water, then filter out the salt." A mixture consists of sand and an aqueous salt solution. The procedure that can be used to separate the sand, salt, and water from each other is to evaporate the water, then filter out the salt.

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Part 1: Name the type of chemical reaction that occurs when magnesium chloride solution (MgCl2) reacts with sodium carbonate sol

garri49 [273]

Answer:

Mg would blow off. AI would be affective to copper but not to MG

Explanation:

7 0

2 years ago

The rate law of the reaction NH3 + HOCl → NH2Cl + H2O is rate = k[NH3][HOCl] with k = 5.1 × 106 L/mol·s at 25°C. The reaction is

SIZIF [17.4K]

Answer:

40% of the ammonia will take 4.97x10^-5 s to react.

Explanation:

The rate is equal to:

R = k*[NH3]*[HOCl] = 5.1x10^6 * [NH3] * 2x10^-3 = 10200 s^-1 * [NH3]

R = k´ * [NH3]

k´ = 10200 s^-1

Because k´ is the psuedo first-order rate constant, we have the following:

b/(b-x) = 100/(100-40) ; 40% ammonia reacts

b/(b-x) = 1.67

log(b/(b-x)) = log(1.67)

log(b/(b-x)) = 0.22

the time will equal to:

t = (2.303/k) * log(b/(b-x)) = (2.303/10200) * (0.22) = 4.97x10^-5 s

6 0

2 years ago

Consider the following system at equilibrium where H° = -87.9 kJ, and Kc = 83.3, at 500 K. PCl3(g) + Cl2(g) PCl5(g) If the VOLUM

7nadin3 [17]

Answer:

The value of Kc C. remains the same.

The value of Qc C. is less than Kc.

The reaction must: A. run in the forward direction to reestablish equilibrium

The number of moles of Cl2 will B. decrease.

Explanation:

Le Chatelier's Principle states that if a system in equilibrium undergoes a change in conditions, it will move to a new position in order to counteract the effect that disturbed it and recover the state of equilibrium.

A decrease in volume causes the system to evolve in the direction in which there is less volume, that is, where the number of gaseous moles is less.

But temperature is the only variable that, in addition to influencing equilibrium, modifies the value of the constant Kc. So if the volume of the equilibrium system is suddenly decreased at constant temperature: The value of Kc remains the same.

As mentioned, if the volume of an equilibrium gas system decreases, the system moves to where there are fewer moles. In this case, being:

PCl₃(g) + Cl₂(g) ⇔ PCl₅(g)

The equilibrium in this case then shifts to the right because there is 1 mole in the term on the right, compared to the two moles on the left. So, The reaction must: A. run in the forward direction to reestablish equilibrium.

By decreasing the volume, and so that Kc remains constant, being:

$Kc=\frac{[PCl_{5} ]}{[PCl_{3}]*[Cl_{2} ]}=\frac{\frac{nPCl_{5} }{Volume} }{\frac{nPCl_{3}}{Volume}*\frac{nCl_{2} }{Volume} } =\frac{nPCl_{5}}{nPCl_{3}*nCl_{2}} *Volume$

where nPCl₅, nPCl₃ and nCl₂ are the moles in equilibrium of PCl₅, PCl₃ and Cl₂

so, the number of moles of Cl₂ should decrease.The number of moles of Cl2 will B. decrease.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system will evolve to the right, the direct reaction prevailing, to increase the concentration of products. So in this case, if the reaction moves to the right, the value of Qc C. is less than Kc.

3 0

2 years ago

Like all equilibrium constants, Kw varies somewhat with temperature. Given that Kw is 3.31 × 10−13 at some temperature, compute

Artyom0805 [142]

Since Kw= [H⁺][OH⁻], and the concentration of both substances are the same, the equation is now Kw=[H⁺]²
So,
3.31x10⁻¹³ = [H⁺]²
Take the square root= 5.75x10⁻⁷
Then take the negative log to find the pH:
-log(5.75x10⁻⁷) = 6.25

5 0

2 years ago

A farmer had an accident and spilled a chemical into his pond. Several days later, he notices many dead frogs around the pond. T

andrezito [222]

B. the frogs are a limiting factor for the gnats
the frogs limit the reproduction of the gnats, and therefore with less frogs the gnat population can increase

3 0

2 years ago

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