In the given chemical equation, [tex]3NaHCO_{3} + CaCl_{2} → 3 NaCl + CaCO_{3}+CO_{2}+H_{2}O[/tex] there are no covalent bonds formed between the elements sodium (Na), hydrogen (H), carbon (C), oxygen (O), calcium (Ca), and chlorine (Cl).
Covalent bonds occur when atoms share electrons to achieve a stable electron configuration. In this equation, the reaction involves the combination of ionic compounds, where ionic bonds are formed between sodium and chlorine ions and between calcium and chloride ions.
These ionic bonds are based on the electrostatic attraction between positively and negatively charged ions. Covalent bonds typically form between nonmetallic elements when they share electrons. In this equation, only sodium, hydrogen, carbon, and oxygen are present as nonmetals.
However, there are no instances where these elements are directly bonded together through shared electrons. Therefore, in the given equation, there are no covalent bonds formed between the elements involved.
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2 A high school student takes a lump of magnesium with a volume of 150.0 mL and adds it to a beaker of
an aqueous solution of aluminum nitrate. What is the mass of the solid aluminum that forms?
Solid magnesium has a density of 1.738 g/cm³.
The mass of the solid aluminum that forms are 192.73 grams
To determine the mass of solid aluminum that forms, we need to use stoichiometry and the balanced chemical equation for the reaction between magnesium and aluminum nitrate.
The balanced chemical equation is:
3 Mg + 2 Al([tex]NO_{3}[/tex])3 → 3 Mg([tex]NO_{3}[/tex])2 + 2 Al
The equation shows that 3 moles of magnesium react with 2 moles of aluminum to produce 2 moles of aluminum nitrate.
To calculate the mass of solid aluminum, we need to know the amount of magnesium used. Given that the volume of the magnesium is 150.0 mL and its density is 1.738 g/cm³, we can calculate the mass of magnesium using the formula:
Mass = Volume × Density
Mass of magnesium = 150.0 mL × 1.738 g/cm³ = 260.7 g
Now, using the molar mass of magnesium (24.31 g/mol) and the molar ratio from the balanced equation, we can determine the moles of magnesium used:
Moles of magnesium = Mass of magnesium / Molar mass of magnesium
= 260.7 g / 24.31 g/mol
= 10.72 mol
According to the stoichiometry of the balanced equation, the ratio of moles of magnesium to moles of aluminum is 3:2. Therefore, the moles of aluminum formed will be:
Moles of aluminum = (2/3) × Moles of magnesium
= (2/3) × 10.72 mol
= 7.15 mol
Finally, we can calculate the mass of solid aluminum using its molar mass (26.98 g/mol):
Mass of aluminum = Moles of aluminum × Molar mass of aluminum
= 7.15 mol × 26.98 g/mol
= 192.73 g
Therefore, the mass of the solid aluminum that forms is approximately 192.73 grams.
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17. HAZWOPER training and certification recognizes:
a. A large number (as much as 80%) will self-present or be self-referred victims
b. Awareness level training will promote proper initial triage actions
c.
Victims will use any entrance they can enter at the hospital, in addition to the
emergency department entrance
d. Both A and C
HAZWOPER training and certification recognize:
a large number (as much as 80%) will self-present or be self-referred victimsVictims will use any entrance they can enter at the hospital, in addition to the emergency department entranceThe correct option is both A and C
What is the HAZWOPER training and certification?HAZWOPER (Hazardous Waste Operations and Emergency Response) training and certification recognize that a large number of victims (as much as 80%) in hazardous waste incidents or emergencies will self-present or be self-referred for medical treatment.
Additionally, HAZWOPER training acknowledges that victims may use any entrance they can access at a hospital, not just the emergency department entrance.
This is because individuals affected by hazardous materials may arrive at different areas of the hospital seeking medical assistance.
Therefore, option d. Both A and C are correct statements regarding the recognition of HAZWOPER training and certification.
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Polyethylene is 86.0% C and 14.0%
H. Determine the empirical formula of the compound.
Percent to Mass: How many grams of C/and Hare present in 100.0 g?
The empirical formula of polyethylene can be determined by converting the given percentages of carbon (C) and hydrogen (H) into grams. To find the grams of each element, we assume a 100.0 g sample of polyethylene.
For carbon:
Mass of carbon = 86.0% × 100.0 g = 86.0 g
For hydrogen:
Mass of hydrogen = 14.0% × 100.0 g = 14.0 g
Therefore, in a 100.0 g sample of polyethylene, there are 86.0 grams of carbon and 14.0 grams of hydrogen.
The empirical formula of a compound represents the simplest whole-number ratio of atoms present in the compound. To determine the empirical formula, we need to find the ratio of carbon to hydrogen in terms of moles.
First, we convert the masses of carbon and hydrogen into moles using their respective molar masses. The molar mass of carbon is approximately 12.01 g/mol, and the molar mass of hydrogen is approximately 1.008 g/mol.
Number of moles of carbon = 86.0 g / 12.01 g/mol ≈ 7.162 mol
Number of moles of hydrogen = 14.0 g / 1.008 g/mol ≈ 13.89 mol
Next, we divide the number of moles of each element by the smallest number of moles to get a simplified ratio.
Carbon: Hydrogen ≈ 7.162 mol : 13.89 mol ≈ 1 : 1.939
Since we want to express the ratio in whole numbers, we multiply both sides by 2 to get a whole number ratio.
Carbon: Hydrogen ≈ 2 : 3.878
Rounding to the nearest whole number, we find that the empirical formula of polyethylene is CH₂.
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A sample dontains 7.90 g C and 42.1 g S. You want to determine the empirical formula.
How many moles of C are in the sample?
P? 1 mol c
Do NOT round your answer.
mol c
The number of moles of carbon (C) in the sample can be determined using its molar mass. The molar mass of carbon is approximately 12.01 g/mol. To calculate the number of moles, we divide the mass of carbon in the sample by its molar mass. In this case, the sample contains 7.90 grams of carbon. Therefore, the number of moles of carbon in the sample is:
7.90 g C / 12.01 g/mol = 0.657 mol C
The molar mass is the mass of one mole of a substance. In this case, the molar mass of carbon is 12.01 g/mol. By dividing the mass of carbon in the sample (7.90 g) by its molar mass, we can calculate the number of moles. This calculation is done because moles are a convenient unit for comparing different elements and compounds in chemical reactions. \The result of 0.657 moles of carbon indicates that there are 0.657 times Avogadro's number (6.022 x 10^23) of carbon atoms in the sample. This information is useful for determining the empirical formula, which represents the simplest whole number ratio of atoms in a compound. To calculate the empirical formula, we would also need to determine the number of moles of sulfur (S) in the sample and find the ratio of the two elements.
In summary, there are approximately 0.657 moles of carbon in the given sample.
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