If the volume of 425 grams was 48.0 cm³, simply divide
g/cm³ = 425 g/48 cm³ = 8.85 g/cm³
If using water in water displacement, 1 mL = 1 cm³
8.85 g/cm³ = 8.85 g/mL
This density is most closely aligned with that of B. Copper
Hope I helped!
<span>Calculating the moles and the moles ratio of the elements gives us the ratio of atoms in each element.
Converting the percentage of element into grams
40.25% carbon = 40.25/100 = .4025 * 100 g of carbon = 40.25g of C
6.19% hydrogen = 6.19/100 = .0619 * 100g g of hydrogen = 6.19g of H
8.94% oxygen = 8.94/100 = .0819 * 100 g of oxygen = 8.19g of O
44.62% bromine = 44.62/100 = .4462 * 100 g of bromine = 44.62g of Br
Converting the grams of element into moles
(48.38 g C) (1 mol/ 12.10 g C) = 4.028 mol C
(8.12 g H) (1 mol/ 1.008 g H) = 8.056 mol H
(53.38 g O) (1 mol/ 16.00 g O) = 3.336 mol O
(44.62g of Br)(0.012515018021626 moles) = 0.55842 mol Br
Calculating the moles ratio of elements by dividing the small number of moles of an element
4.028 mol C /0.55842 = 7.2 mol C x 5 = 36 mol C
8.056 mol H / 0.55842 = 14.42 mol H = 72 mol H
3.336 mol O / 0.55842 = 5.97 mol O = 30 mol O
0.55842 mol Br / 0.55842 = 1mol Br = 5 mol Br
So the empirical formula is (C6H12O5)6Br5</span>
Answer:
Energy transfers from the metal to the water and calorimeter until they are all at room temperature.
Explanation:
CHECK THE COMPLETE QUESTION BELOW;
A metal sample is heated and placed into the water in a calorimeter at room temperature. Which statement best describes how the calorimeter can be used to determine the specific heat capacity of the metal sample?
Energy transfers to the metal from the water and calorimeter until they are all at room temperature
. Energy transfers from the metal to the water and calorimeter until they are all at room temperature.
Energy transfers to the metal from the water and calorimeter until they all reach a single temperature.
Energy transfers from the metal to the water and calorimeter until they all reach a single temperature.
EXPLANATION;
Using calorimeter to determine the specific heat capacity of the metal sample can be associated to the theory of conservation of energy because heat which is a form of energy is been transfer of heat between the metal to the water and the calorimeter, this process will proceed till single temperature is attained.
The change in the amount of temperature of the water in the calorimeter is measured in order to get the difference in heat change of the calorimeter water.
CHANGE IN HEAT CAN BE CALCULATED USING THE FORMULA.
Q = cmΔT where Q is the change in heat , c is the specific heat capacity and ΔT is the change in temperature
As the pressure on the on a gas cofined above a liquid increases, the solubility of the gas will increase
this also happen when we lower the temperature
MgBr2, 3 ions per mole=best conductor
KBr, 2 ions per mole, 2nd best conductor
Benzoic acid, weak acid, slightly ionized, weak conductor
Sugar, molecular, non ionized, non conductor
