When utilizing the gravimetric method, it is crucial to completely dissolve your sample in 10 mL of water. A quantitative technique called gravimetric analysis employs the selective precipitation of the component under study from an aqueous solution.
A group of techniques known as gravimetric analysis are employed in analytical chemistry to quantify an analyte based on its mass. Gravimetric analysis is a quantitative chemical analysis technique that transforms the desired ingredient into a substance (of known composition) that can be extracted from the sample and weighed. This is a crucial point to remember.
Gravimetric water content (g) is therefore defined as the mass of water per mass of dry soil. To calculate it, weigh a sample of wet soil, dry it to remove the water, and then weigh the dried soil (mdry). Dimensions of the sample Water is commonly forgotten despite having a density close to one.
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Answer: The fourth material that is added to the blast furnace is HOT AIR which provides OXYGEN for used for combustion of carbon (Coke).
Explanation:
Iron is the second most abundant metal found in the earth's crust after aluminium. It is not found in the free metallic state but are extracted from rocks which are rich in iron that contains other materials. These are known are iron ores and the most common iron ores are haematite ( Fe2O3).
Iron can be extracted from its ore with the used of blast furnace. The materials used for extraction of iron includes:
--> Coke
--> haematite( iron ore)
--> limestone and
--> Hot air.
The iron ore is first roasted in air so that iron(III)oxide is produced. The iron(III)oxide is then mixed with coke and limestone and heated to a very high temperature. Hot air is introduced into it from the bottom of the furnace. The coke is oxidizes the the oxygen in the hot air blast to liberate carbondioxide.
132.13952<span> grams - I believe</span>
Answer:
The answers are in the explanation.
Explanation:
The energy required to convert 10g of ice at -10°C to water vapor at 120°C is obtained per stages as follows:
Increasing temperature of ice from -10°C - 0°C:
Q = S*ΔT*m
Q is energy, S specific heat of ice = 2.06J/g°C, ΔT is change in temperature = 0°C - -10°C = 10°C and m is mass of ice = 10g
Q = 2.06J/g°C*10°C*10g
Q = 206J
Change from solid to liquid:
The heat of fusion of water is 333.55J/g. That means 1g of ice requires 333.55J to be converted in liquid. 10g requires:
Q = 333.55J/g*10g
Q = 3335.5J
Increasing temperature of liquid water from 0°C - 100°C:
Q = S*ΔT*m
Q is energy, S specific heat of ice = 4.18J/g°C, ΔT is change in temperature = 100°C - 0°C = 100°C and m is mass of water = 10g
Q = 4.18J/g°C*100°C*10g
Q = 4180J
Change from liquid to gas:
The heat of vaporization of water is 2260J/g. That means 1g of liquid water requires 2260J to be converted in gas. 10g requires:
Q = 2260J/g*10g
Q = 22600J
Increasing temperature of gas water from 100°C - 120°C:
Q = S*ΔT*m
Q is energy, S specific heat of gaseous water = 1.87J/g°C, ΔT is change in temperature = 20°C and m is mass of water = 10g
Q = 1.87J/g°C*20°C*10g
Q = 374J
Total Energy:
206J + 3335.5 J + 4180J + 22600J + 374J =
30695.5J =
30.7kJ
To obey the Law of Conservation of Mass, the sum of all individual elements of a compound is equal to the mass of the compound. So, if HCN has a mass of 7.83 grams, then
7.83 g = mass of H + mass of C + mass of N
We know the masses of H and N to be 0.290 g and 4.06 g, respectively. Hence, we can find for the mass of C:
7.83 = 0.29 + mass of C + 4.06
mass of C = 3.48 g
As an extension to the Law of Conservation of Mass, there is also a Law of Definite Proportions. According to Dalton's atomic theory, a compound is formed from a fixed ratio of its individual elements. From our previous calculations, we know that the mass ratio of H to C to N is 0.29 g: 3.48 g:4.06 grams. The ratio could also be expressed in percentages. Let's find the mass percentage of Carbon in HCN to be used later:
mass % of Carbon = (3.48 g/7.83 g)*100
mass % of Carbon = 44.44%
So, if you collect a different mass of HCN, say 3.37 g, the corresponding mass of Carbon is equal to:
Mass of Carbon = (3.37)(44.44%)
Mass of Carbon = 1.498 g
