Where are metals on the table?

A 12.0 g sample of a metal is heated to 90.0 ◦C. It is then dropped into 25.0 g of water. The temperature of the water rises fro

deff fn [24]

Answer:

The specific heat of the metal is 0.34 J/g°C

Explanation:

Step 1: Data given

Mass of the metal = 12.0 grams

Mass of the water = 25.0 grams

Initial temperature of the metal = 90.0 °C

Initial temperature of the water = 22.5 °C

Final temperature = 25 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of the metal

Qlost = Qgained

Q = m*c*ΔT

Qmetal = -Qwater

m(metal) *c(metal)* ΔT(metal) = -m(water) * c(water) *ΔT(water)

⇒ with mass of metal = 12.0 grams

⇒ with c(metal) = TO BE DETERMINED

⇒ with ΔT(metal) = T2 - T1 = 25.0°C - 90.0 °C = -65.0 °C

⇒ with mass of water = 25.0 grams

⇒ with c(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2 - T1 = 25.0 - 22.5 °C = 2.5 °C

12.0 * c(metal) * -65.0 °C = -25.0g * 4.184 J/g°C * 2.5°C

-780.0 * c(metal) = -2615 ( 2.6*10^3 with sig figs)

c(metal) = 0.335 (=0.34 with sig figs)

The specific heat of the metal is 0.34 J/g°C

3 0

3 years ago

2. Which statement is TRUE about ionic compounds?

oksian1 [2.3K]

Answer:

I think it's c I hope this helps

6 0

2 years ago

What would happen if we did not have an atmosphere?

olganol [36]

Answer:

The air would still be too thin to breathe. The lack of atmosphere would chill the Earth's surface. Organisms that need air to breathe would die.

Explanation:

Hope it helped!

6 0

3 years ago

Concept of mole chemistry question

vredina [299]

Answer:

How many molecules of water are there in 54 g of H2O H 2 O ? Solution. Molar Mass of H2O H 2 O = 2 + 16 = 18 g/moles. So ,number of moles of H2O H 2 O = Mass/Molar Mass = 54/18 =3 moles. Now 1 moles = 6.022×1023 6.022 × 10 23 molecule.

5 0

2 years ago

Enter your answer in the provided box. phosphorus pentachloride is used in the industrial preparation of organic phosphorus comp

ankoles [38]

I have a helpful trick for such problems.

Step 1:
Write the equation asked with exact moles given in question,

PCl₃ + Cl₂ → PCl₅ --------(1)
Step 2:
Write equations given in data along with energies,

P₄ + 6Cl₂ → 4PCl₃ ΔH = −1280 kj -----(2)

P₄ + 10Cl₂ → 4PCl₅ ΔH = −1774 kj -----(3)

Step 3:
Compare the equation 1 with eq 2 and 3 and modify eq 2 and 3 according to eq 1 both in terms of reactant and product and number of moles. For example in eq 1 PCl₃ is on left hand side and its number of moles is one, so, modify eq, 2 according to eq 1 both in terms of moles and reactant product sides. i.e.
By inter converting eq 2, the sign of energy will change from negative to positive, and divide eq 2 with 4, to have its ratio's equal to eq 1

PCl₃ → 1/4 P₄ + 1.5 Cl₂ ΔH = +320 KJ

Same steps are done for eq 3, but in this case signs are not changed only moles are changed

1/4P₄ + 2.5Cl₂ → 1/4PCl₅ ΔH = −443.5 kj
Now,
Compare both last equations,

PCl₃ → 1/4 P₄ + 1.5 Cl₂ ΔH = +320 KJ

1/4P₄ + 2.5Cl₂ → 4/4PCl₅ ΔH = −443.5 kj
_____________________________________________

PCl₃ + Cl₂ → PCl₅ ΔH = -123.5 KJ

7 0

3 years ago