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3 years ago

If a reaction occurs when a piece of metal is placed in a solution, what can you conclude about the solution?

1 answer:

4 0

Hello!

If a reaction occurs when a piece of metal is placed in a solution, you can conclude that the solution is probably acidic because bases rarely react with metals.

Strong Acids, like HCl, react with metals to produce salts and release gaseous hydrogen (H₂) which is evidenced by the generation of bubbles in the solution. The general chemical equation for this kind of reactions for a metal M and an acid HA is:

2M(s) + 2HA(aq) → 2MA(aq) + H₂(g)

Have a nice day!

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Which of the following describes experiences initiated spontaneously by children as they go about their daily activities?

Harlamova29_29 [7]

Answer:

naturalistic

Explanation:

4 0

3 years ago

Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f

N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

Answer: The initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

Explanation:

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+2B\rightleftharpoons C$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first trial:

$5.4\times 10^{-3}=k(0.30)^a(0.050)^b$ ....(1)

Expression for rate law for second trial:

$1.1\times 10^{-2}=k(0.30)^a(0.100)^b$ ....(2)

Expression for rate law for third trial:

$2.2\times 10^{-2}=k(0.50)^a(0.050)^b$ ....(3)

Dividing 2 by 1, we get:

$\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1$

Dividing 3 by 1, we get:

$\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^1$ ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}$

Calculating the initial rate of formation of C by using equation 4, we get:

$k=1.2M^{-2}s^{-1}$

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

$\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}$

Hence, the initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

8 0

3 years ago

How many grams of sodium ions are in na2so4?

LiRa [457]

Answer:

142.04

Explanation:

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3 years ago

A truck traveled at a constant speed for 6 hours. Then, it

gavmur [86]

Answer:

2.4 hrs

Explanation:

The constant speed of the truck for 6 hrs can be calculated by: speed=distance/time. Speed =(876-228)/6=648/6=108m/s. So the decreased speed = (108-13)=95m/h. Now, speed =distance /time We get 95m/h = 228/t. t=228/95 hrs = 2.4 hrs PLEASE MARK ME THE BRAINLIEST!!

7 0

3 years ago

Read 2 more answers

Solid iron metal reacts with solid sulfur at room temperature to produce solid iron (ll) sulfide? Express as a chemical equation

kupik [55]

Answer:Fe+S=FeS

Hope this helps

5 0

2 years ago