Answer:
Mass of CO2 produced = 944.43 -lb/year
Since 944.43 -lb is close to 1000 -lb, it is a reasonable statement.
Explanation:
Assumptions:
Car mileage = 20 miles/gallon:
Density of gasoline = 0.702 g/ml
Volume of gallons per week = (20 miles/week) / (20 miles/gallon) = 1 gallon/week
Volume of gasoline per year = 52 weeks * 1 gallon/week = 52 gallons
1 gallon = 3.8 litres or 3800 ml
52 gallons = 52 * 3800 = 197600 ml
Mass of gasoline = volume *density
Mass of gasoline = 197600 ml * 0.702 g/ml
Mass of gasoline = 138715.2 g
Equation of reaction:
2C8H18 + 25O2 ---> 16CO2 + 18H2O
2 moles of octane produces 16 moles of CO2
Molar mass of octane = 114 g/mol; molar mass of CO2 = 44 g/mol
114 * 2 g of octane = 44 * 16 g of CO2
138715. 2 g of octane produces = (138715.2 * 44 * 16) / 114 * 2 of CO2
Mass of CO2 produced = 428313.6 g of CO2 = 428.3136 Kg
Mass of CO2 produced in -lb weight = 428.3136 * 2.205
Mass of CO2 produced per year = 944.43 -lb CO2
Since 944.43-lb is close to 1000 -lb, therefore, it is a reasonable statement.
Answer:
freezing point (°C) of the solution = - 3.34° C
Explanation:
From the given information:
The freezing point (°C) of a solution can be prepared by using the formula:
where;
i = vant Hoff factor
the vant Hoff factor is the totality of the number of ions in the solution
Since there are 1 calcium ion and 2 nitrate ions present in Ca(NO3)2, the vant Hoff factor = 3
= 1.86 °C/m
m = molality of the solution and it can be determined by using the formula
which can now be re-written as :
molality = 0.599 m
∴
The freezing point (°C) of a solution can be prepared by using the formula:
the freezing point of water - freezing point of the solution
3.34° C = 0° C - freezing point of the solution
freezing point (°C) of the solution = 0° C - 3.34° C
freezing point (°C) of the solution = - 3.34° C
Answer and Explanation :
Paramagnetic are those which has unpaired electrons and diamagnetic are those in which all electrons are paired.
(a) V³⁺
The electronic configuration is -
The electrons in 3d orbital = 1 (Unpaired)
Thus, the ion is paramagnetic as the electrons are unpaired.
(b) Cd²⁺
The electronic configuration is -
The electrons in 4d orbital = 10 (paired)
Thus, the ion is diamagnetic as the electrons are paired.
(c) Co³⁺
The electronic configuration is -
The electrons in 3d orbital = 6 (Unpaired)
Thus, the ion is paramagnetic as the electrons are unpaired.
(d) Ag⁺
The electronic configuration is -
The electrons in 4d orbital = 10 (paired)
Thus, the ion is diamagnetic as the electrons are paired.
Answer:
The correct answer is option B.
Explanation:
Significant figures :These are figures in a number which expresses the value of the magnitude of a quantity to a specific degree of accuracy is known as significant digits.
Significant digits in the volume measurement 0.010 mL are:
= 2 significant figures
The volume measurement "0.010 mL" contains two significant digits. Significant digits are nonzero digits and trailing zeros in decimal numbers that contribute to the precision of the measurement.
In the volume measurement "0.010 mL," there are two significant digits. Significant digits are the digits in a measured quantity that contribute to its precision or accuracy. To identify significant digits, consider the following rules:
1. All nonzero digits are significant. In this case, "0" is considered a nonzero digit because it appears between two nonzero digits.
2. Leading zeros (zeros to the left of the first nonzero digit) are not significant. In "0.010 mL," the leading zero is not considered a significant digit.
3. Trailing zeros (zeros to the right of the last nonzero digit) in a decimal number are significant. So, the two trailing zeros in "0.010 mL" are significant.
Therefore, the measurement "0.010 mL" has two significant digits (the "1" and the "0"), and the correct answer is B) 2. These two significant digits represent the precision of the volume measurement.
Learn more about significant digits here:
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Answer:
(0,653±0,002) M of HNO₃
Explanation:
The reaction of standarization of HNO₃ with Na₂CO₃ is:
2 HNO₃ + Na₂CO₃ ⇒ 2 Na⁺ + H₂O + CO₂ + 2NO₃⁻
To obtain molarity of HNO₃ we need to know both moles and volume of this acid. The volume is (27,71±0,05) mL and to calculate the moles it is necessary to obtain the Na₂CO₃ moles and then convert these to HNO₃ moles, thus:
0,9585 g of Na₂CO₃ × ( 1 mole / 105,988 g) =
9,043×10⁻³ mol Na₂CO₃ × ( 2 moles of HNO₃ / 1 mole of Na₂CO₃) = 1,809×10⁻² moles of HNO₃
Molarity is moles divide liters, thus, molarity of HNO₃ is:
1,809×10⁻² moles / 0,02771 L = 0,6527 M of HNO₃
The absolute uncertainty of multiplication is the sum of relative uncertainty, thus:
ΔM = 0,6527M× (0,0007/0,9585 + 0,001/105,988 + 0,05/27,71) =
0,6527 M× 2,54×10⁻³ = 1,7×10⁻³ M
Thus, molarity of HNO₃ solution and its absolute uncertainty is:
(0,653±0,002) M of HNO₃
I hope it helps!
Answer with double replacement
Answer:
Barium chloride + Sodium phosphate → barium phosphate + sodium chloride
Explanation:
Double replacement:
It is the reaction in which two compound exchange their ions and form new compounds.
AB + CD → AC +BD
Chemical equation:
BaCl₂ + Na₃PO₄ → Ba₃(PO₄)₂ + NaCl
Balanced chemical equation:
3BaCl₂ + 2Na₃PO₄ → Ba₃(PO₄)₂ + 6NaCl
The cation and anion of both reactants are exchanged with each other.
Ba²⁺ react with PO₄³⁻ and form Ba₃(PO₄)₂ while Cl⁻ react with Na⁺ and form sodium chloride.
Molecular equation:
Barium chloride + Sodium phosphate → barium phosphate + sodium chloride
A double replacement reaction between barium chloride and sodium phosphate results in the formation of barium phosphate and sodium chloride.
The given chemical equation represents a double replacement reaction. The chemical reactants are barium chloride (BaCl2) and sodium phosphate (Na3PO4). In a double replacement reaction, the cations and anions of the two reactants switch places to form two new compounds.
So here is how the reaction would proceed: BaCl2 + Na3PO4 -> Ba3(PO4)2 + NaCl.
This translates to: Barium chloride reacts with sodium phosphate to form barium phosphate and sodium chloride.
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Answer:
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