The density of the iron pyrite cube is 1.343 g/cm³.
Given,
Side of iron pyrite cube = 0.31 cm
Mass of iron pyrite = 0.040 g
The volume of iron pyrite cube = s³ cm³
Or, volume = 0.029791 cm³
We have to find the density of the sample.
Density is defined as the mass per unit volume. Or, it is the ratio of mass to the volume of the substance.
Using the formula for density, we get,
Density = mass/volume
Or, density = 0.40/0.029791
Or, density = 1.343 g/cm³
Hence, the density of the iron pyrite cube is 1.343 g/cm³.
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The volume of ideal gas that would occupy at 1.50 atm and 216 c is
2016.49 ml
<u><em>calculation</em></u>
- volume is calculated using a combined gas equation
that is P1V1/T1= P2V2/T2
P1( pressure 1) = 7.20 atm
V1 ( volume 1) = 250 ml
T1 ( temperature 1) = 18 + 273=291 K
P2 ( pressure 2) = 1.50 atm
T2 ( temperature 2)= 216+273=489 K
V2 ( volume 2) =?
- by making V2 the subject of the formula
V2=[(T2 xP1 x V1)/(T1 xP2)]
V2=[ (489 K x 7.20 atm x 250 ml) / ( 291 k x 1.50 atm)] = 2016.49 Ml
Answer:
0.5 × 10²³ atoms of iodine
Explanation:
Given data:
Mass of calcium iodide = 12.75 g
Number of atoms of iodine = ?
Solution:
First of all we will calculate the number of moles of calcium iodide.
Number of moles = mass/ molar mass
Number of moles = 12.75 g/ 293.9 g/mol
Number of moles = 0.04 mol
In one mole of calcium iodide there are two moles of iodine.
Thus in 0.04 moles:
0.04 mol × 2 = 0.08 moles of iodine
Now we will use the Avogadro number:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ atoms
0.08 moles of iodine × 6.022 × 10²³ atoms / 1 mol
0.5 × 10²³ atoms of iodine.
Answer:
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It is a laboratory-based science that brings together
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Answer:
Q = 360 Joules
Explanation:
specific heat of copper = 0.400 J/g° C
mass of copper = 30 g
initial temperature = 20.0°C
final temperature = 50.0°C
Using the formula:
Q = mcΔT
where;
Q = Heat Energy
Q = (30 × 0.400)(50-20)
Q = 360 Joules