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My name is Ann [436]
3 years ago
14

60 grams of ice will require _____ calories to raise the temperature 1°C.

Chemistry
1 answer:
guajiro [1.7K]3 years ago
7 0

Answer:

60 grams of ice will require 30.26 calories to raise the temperature 1°C.

Explanation:

The amount of heat (Q) to raise the temperature of 60.0 g of ice by 1°C can be calculated from:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat released or absorbed by the system.

m is the mass of the ice (m = 60.0 g).

c is the specific heat capacity of ice (c = 2.108 J/g.°C).

ΔT is the temperature difference (ΔT = 1.0 °C).

∴ Q = m.c.ΔT = (60.0 g)(2.108 J/g.°C)(1.0 °C) = 126.48 J.

<em>It is known that 1.0 cal = 4.18 J.</em>

<em>∴ Q = (126.48 J)(1.0 cal / 4.18 J) = 30.26 cal.</em>

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Write the appropriate symbol for each of the following isotopes: (a) Z 11, A 23; (b) Z= 28, A= 64; (c) Z= 50, A =115; (d) Z= 20,
prohojiy [21]

Explanation:  

Z = atomic mass of the element and  , A = atomic mass of the element .

a) Z = 11, A =  23

Element = Sodium

  symbol: ²³₁₁Na  .

b) Z = 28, A =  64

Element = Nickel

  symbol: ⁶⁴₂₈Ni  .

c) Z = 50, A = 115

Element = tin

  symbol: ¹¹⁵₅₀Sn  .

d) Z = 20, A = 42

Element = Calcium

  symbol: ⁴²₂₀Ca .

6 0
3 years ago
Read 2 more answers
How many molecules are in 1kg of water
Mila [183]

Answer:

334.2× 10²³ molecules

Explanation:

Given data:

Mass of water = 1 Kg ( 1000 g )

Number of molecules = ?

Solution:

Number of moles of water:

Number of moles = mass/ molar mass

Number of moles = 1000 g/ 18 g/mol

Number of moles = 55.5 mol

1 mole contain 6.022× 10²³ molecules

55.5 mol×6.022× 10²³ molecules

334.2× 10²³ molecules

8 0
3 years ago
Anabelle has noticed that her strawberry plants haven't produced any strawberries yet this year.
marusya05 [52]

Answer:0.260

Explanation:

4 0
3 years ago
What is melted rock and minerals found beneath earths rust called
satela [25.4K]
Magma- hot fluid or semifluid material below or within the earth's crust from which lava and other igneous rock is formed by cooling. Hope this helps!
3 0
2 years ago
For a particular isomer of C 8 H 18 , C8H18, the combustion reaction produces 5104.1 kJ 5104.1 kJ of heat per mole of C 8 H 18 (
vlada-n [284]

Answer:

The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol

Explanation:

Step 1: Data given

The combustion reaction of octane produces 5104.1 kJ per mol octane

Step 2: The balanced equation

C8H18(g) + 12.5 O2 ⟶ 8CO2 (g) + 9 H2O (g)  ∆H°rxn = -5104.1 kJ/mol

Step 3:

∆H°rxn = ∆H°f of products minus the ∆H° of reactants

∆H°rxn = ∆H°f products - [∆H°f reactants]

-5104.1 kJ/mol = (8*∆H°fCO2 + 9*∆H°fH20) - (∆H°fC8H18 + 12.5∆H°fO2)

∆H°f C8H18 = ∆H°f 8CO2 + ∆H°f 9H2O+ 5104.1 kJ/mol

∆H°f C8H18 = 8 * (-393.5 kJ)/mol + 9 * (-241.8 kJ/mol)] + 5104.1 kJ /mol

∆H°f C8H18 = -220.1 kJ/mol

The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol

7 0
3 years ago
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