The density of the metal object=6.0
Given:
Volume of the metal object=1.5ml
Mass of the metal object=9.0g
To find:
Density of the metal object
<u>Step by Step Explanation:
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Solution:
According to the formula, Density of the metal object can be calculated as
Where, m=mass of the metal object
=density of the metal object
v=volume of the metal object
We know the values of v=1.5ml and m=9.0g
Substitute these values in the above equation we get
=9.0/1.5
=6.0
Result:
Thus the density of the metal object is 6.0
Because of the two textures aren't supposed to be washed together. Wash and dry them seperatly then you won't have the problem. They cause friction.
Answer:
It would take 3.11 J to warm 3.11 grams of gold
Explanation:
Step 1: Data given
Mass of gold = 3.11 grams
Temperature rise = 7.7 °C
Specific heat capacity of gold = 0.130 J/g°C
Step 2: Calculate the amount of energy
Q = m*c*ΔT
⇒ Q = the energy required (in Joules) = TO BE DETERMINED
⇒ m = the mass of gold = 3.11 grams
⇒ c = the specific heat of gold = 0.130 J/g°C
⇒ ΔT = The temperature rise = 7.7 °C
Q = 3.11 g * 0.130 J/g°C * 7.7 °C
Q = 3.11 J
It would take 3.11 J to warm 3.11 grams of gold
Answer: 0.811m
Explanation: Molarity is the moles of solute present per kg of solvent.
Molar mass of C10H8 = . 128.17 g/mol
No. of moles = given mass/molar mass
11.2g C10H8 = 11.2/128.17 g/mol C10H8
= 0.0874 mol C10H8
NOW 107.8g solvent (chloroform) contains 0.0879 molecules (naphthalene)
1000g '' will contain (0.0879/107.8 x 1000) mole solute
= 0.811 mole
Molarity of the solution is 0.811 m (option A)
Answer:
Heat transfer = 3564 Jolues
The same value
Explanation:
The heat of combustion is the heat released per 1 mole of substunce experimenting the combustion at standard conditions of pressure and temperature ( 1 atm, 298 K):
Qtransfer = - mol x ΔHºc Qtransfer
So look up in appropiate reference table ΔHºc and solve the problem:
ΔHºc = - 891 kJ/mol
Qtransfer = - (4 x 10³ mol x -891 kJ/mol ) = 3564 J
if the combustion were achieved with 100 % excess air, the result will still be the same. As long as the standard conditions are maintained, the heat of combustion remains constant. In fact in many cases the combustion is performed under excess oxygen to ensure complete combustion.
