Answer:
The powder increased the hydrogen ion concentration in the water
Explanation:
The ion that is responsible for acidity is the hydrogen ion. This is why all acids have the hydrogen ion in them. Examples are HCl, HBr, HNO3, H2SO4. So a solution with a higher concentrations of hydrogen ions in solution will be more acidic than one with less hydrogen ions in solution, So clearly in this case, the reason this powder would change the pH of water from 7 to 2.7 would be because there has been an increase in acidity of the solution or an increase in the concentration of hydrogen ions
Answer: 61%
The reaction equation should be
CaF2 + H2SO4 → 2HF + CaSO4
For every 1 molecule CaF2 used, there will be 2 molecules of HF formed. The molecular mass of CaF2 is 78/mol while the molecular mass of HF is 20g/mol. If the yield is 100%, the amount of HF formed by 112g CaF2 would be: 112g/(78g/mol) * 2 * (20g/mol)=57.43g
The percentage yield of the reaction would be: 35g/57.43g= 60.94%
B. density
C. volume
D. temperature
The number of kilograms of fuel that are contained in one canister is 1.03 Kilograms
calculation
mass = density x volume
Density= 0.710 g/cm³
volume = 1.45 L
convert 1.45 l into cm³ to make the units uniform
that is 1 L= 1000 cm³
1.45 L= cm³
by cross multiplication
= (1.45 L x 1000 cm³) / 1 L = 1450 cm³
mass is therefore = 0.710g/cm³ x 1450 cm³ =1029.5 grams
convert grams to Kg
1 kg = 1000 grams
? kg =1029.5 grams
by cross multiplication
= (1 kg x 1029.5 grams) /1000 grams = 1.0295 Kg ≅1.03 kg
the canister contains a gas used to fuel a burner
the volume of the white gas to be used is - 1.45 L
density of white gas is 0.710 g/cm³
density is defined as the mass per unit volume
the mass of 1 cm³ of white gas is - 0.710 g
1 cm³ is equivalent to 0.001 L
the volume given is in Litres
so if 0.001 L has a mass of 0.710 g
then 1.45 L have a mass of - 0.710 g / 0.001 L x 1.45 L = 1030 g
since 1000 g is equivalent to 1 kg
mass of white fuel is - 1030 g/1000 g/kg = 1.03 kg
theres 1.03 kg of fuel in one canister
The best name for the compound is DINITROGEN TRIOXIDE.
Each chemical compound is always represented by a chemical symbol, which shows the ratio at which each atom of the elements of the compound are combine together and this is often used in naming the compound. Looking at the compound given in the question, the compound is made up of two atoms of nitrogen and three atoms of oxygen and this fact was used in naming the compound. In naming chemical compounds, 'Di' stands for 2 while 'Tri' stands for 3. Since there are two nitrogen and three oxygen atoms in the compound, that was why it was named dinitrogen trioxide.
(2) 2 (4) 20
Answer: industrial smog.
Explanation:
The industrial smog typically exists in the urban industrial areas where the large scale burning of the fossil fuels may lead to generation of huge amount of smoke and sulfur dioxide gas liberated into the atmosphere. These gaseous products gets mixed with the droplets of fog already present in the atmosphere. This lead to the formation of industrial smog. The industrial smog forms a thick blanket of smog or haze nearly close to the ground.
Sulfur dioxide is a primary component of industrial smog and photochemical smog, but it does not directly influence thermal inversion. It is released into the atmosphere predominantly from burning fossil fuels.
Sulfur dioxide is a significant component in the creation of both industrial smog and photochemical smog, two serious urban air pollutants. During industrial activities, fossil fuels such as coal and oil that contain sulfur are burned, releasing sulfur dioxide into the atmosphere. This gas then reacts with water vapor and other compounds to form industrial smog. In the case of photochemical smog, when sulfur dioxide interacts with sunlight, volatile organic compounds, and nitrogen oxides, it contributes to the formation of this harmful type of smog. However, sulfur dioxide does not directly relate to the occurrence of thermal inversion, a weather-related phenomenon that can trap smog and other pollutants close to the ground.
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