The process of blank is when two atoms combine together form a larger atom releasing energy

A 13.9 - g piece of metal ( specific heat capacity is 0.449 /g^ C)who whose temperature is 54.2 degrees * C was added to a sampl

lubasha [3.4K]

Answer:

26.2g = Mass of water in the calorimeter

Explanation:

The heat absorbed for the water is equal to the heat released for the metal. Based on the equation:

Q = m*C*ΔT

Where Q is heat, m is the mass of the sample, C is specific heat of the material and ΔT is change in temperature

Replacing we can write:

$m_{metal}*C_{metal}*dT_{metal}=m_{water}*C_{water}*dT_{water}$

13.9g * 0.449J/g°C * (54.2°C-15.6°C) = m(H₂O) * 4.184J/g°C * (15.6°C-13.4°C)

240.9J = m(H₂O) * 9.2J/g

<h3>26.2g = Mass of water in the calorimeter</h3>

4 0

3 years ago

A particular reactant decomposes with a half‑life of 113 s when its initial concentration is 0.331 M. The same reactant decompos

algol13

Answer:

The reaction is second-order, and k = 0.0267 L mol^-1 s^-1

Explanation:

Step 1: Data given

The initial concentration is 0.331 M

half‑life time = 113 s

The same reactant decomposes with a half‑life of 243 s when its initial concentration is 0.154 M.

Step 2: Determine the order

The reaction is not first-order because the half-life of a first-order reaction is independent of the initial concentration:

t½ = (ln(2))/k

Calculate k for the two conditions given:

⇒ 113 s with initial concentration is 0.331 M

t½ = ([A]0)/2k

113 s = (0.331 M)/2k

k = 0.00146 mol L^-1 s^-1

⇒ 243 s with an initial concentration is 0.154 M

t½ = ([A]0)/2k

243 s = (0.154 M)/2k

k = 0.000317 mol L^-1 s^-1

The values of k are different, so that rules out zero-order.

Step 3: Calculate if it's a second-order reaction

For a second-order reaction, the half-life is given by the expression

t½ = 1/((k*)[A]0))

Calculate k for the two conditions given:

⇒ 113 s when its initial concentration is 0.331 M

t½ = 1/((k*)[A]0))

113 s = 1/(k*(0.331 M))

k = 1/((0.331 M)*(113 s)) = 0.0267 L mol^-1 s^-1

⇒ 243 s when its initial concentration is 0.154 M

t½ = 1/((k*)[A]0))

243 s = 1/(k*(0.154 M))

k = 1/((0.154 M)*(243 s)) = 0.0267 L mol^-1 s^-1

The values of k are the same, so the reaction is second-order, and k = 0.0267 L mol^-1 s^-1

4 0

4 years ago

How much excess reactant is left over when 17.0 g of potassium hydroxide (KOH) reacts with

dolphi86 [110]

Answer:

4.56 g of KOH

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2KOH + Fe(NO₃)₂ —> Fe(OH)₂ + 2KNO₃

Next, we shall determine the masses of KOH and Fe(NO₃)₂ that reacted from the balanced equation. This is can be obtained as:

Molar mass of KOH = 39 + 16 + 1 = 56 g/mol

Mass of KOH from the balanced equation = 2 × 56 = 112 g

Molar mass of Fe(NO₃)₂ = 56 + 2[14 + (16×3)]

= 56 + 2[14 + 48)]

= 56 + 2[62]

= 56 + 124

= 180 g/mol

Mass of Fe(NO₃)₂ from the balanced equation = 1 × 180 = 180 g

SUMMARY:

From the balanced equation above,

112 g of KOH reacted with 180 g of Fe(NO₃)₂

Next, we shall determine the limiting reactant and the excess reactant. This can be obtained as follow:

From the balanced equation above,

112 g of KOH reacted with 180 g of Fe(NO₃)₂.

Therefore, 17 g of KOH will react with = (17 × 180)/112 = 27.32 g of Fe(NO₃)₂

From the calculations made above, we can see that it will take a higher mass (i.e 27.32 g) of Fe(NO₃)₂ than what was given (i.e 20 g) to react completely with 17 g of KOH.

Therefore, Fe(NO₃)₂ is the limiting reactant and KOH is the excess reactant.

Next, we shall determine the mass of the excess reactant that reacted. This can be obtained as follow:

From the balanced equation above,

112 g of KOH reacted with 180 g of Fe(NO₃)₂.

Therefore Xg of KOH will react with 20 g of Fe(NO₃)₂ i.e

Xg of KOH = (112 × 20)/180

Xg of KOH = 12.44 g

Thus, 12.44 g of KOH reacted.

Finally, we shall determine the leftover mass of the excess reactant.

The excess reactant is KOH. The leftover mass can be obtained as follow:

Mass of KOH given = 17 g

Mass of KOH that reacted = 12.44 g

Mass of KOH leftover =?

Mass of KOH leftover = (Mass of KOH given) – (Mass of KOH that reacted)

Mass of KOH leftover = 17 – 12.44

Mass of KOH leftover = 4.56 g

Thus, the excess reactant (i.e KOH) that is left over is 4.56 g

3 0

3 years ago

Which of the following statements best explains why atoms bond?

Keith_Richards [23]

Answer:

atoms bond to attain neutral configuration or to get noble gas configuration.

Explanation:

atoms always want to be stable by attaining noble gas configuration.

they do so by forming bonds. they also bond to each other to make their outer electrons stable that is if the atom has an excessive electron than its neutral configuration it can bond to another atom that has one less electron than its neutral configuration by ionic bond.

7 0

3 years ago

Does anyone know the answer to this

Damm [24]

It’s Tetracarbon dioxide :)

6 0

3 years ago