Answer:
1.7 × 10⁴ J
Explanation:
Step 1: Calculate the heat required to raise the temperature of ice from -15 °C to 0°C
We will use the following expression.
Q₁ = c(ice) × m × ΔT
Q₁ = 2.03 J/g.°C × 25 g × [0°C - (-15°C)] = 7.6 × 10² J
Step 2: Calculate the heat required to melt 25 g of ice
We will use the following expression.
Q₂ = C(fusion) × m
Q₂ = 80. cal/g × 25 g × 4.184 J/1 cal = 8.4 × 10³ J
Step 3: Calculate the heat required to raise the temperature of water from 0°C to 75 °C
We will use the following expression.
Q₃ = c(water) × m × ΔT
Q₃ = 4.184 J/g.°C × 25 g × (75°C - 0°C) = 7.8 × 10³ J
Step 4: Calculate the total heat required
Q = Q₁ + Q₂ + Q₃
Q = 7.6 × 10² J + 8.4 × 10³ J + 7.8 × 10³ J = 1.7 × 10⁴ J
(B) False
Isopropyl methyl ether is slightly soluble in water because the oxygen atom of ethers with 3 or lesser carbon atoms can form hydrogen bonds with water. Therefore, the given statement is true.
Hydrogen bonding is a special class of attractive intermolecular forces that arise because of the dipole-dipole interaction between hydrogen that is bonded to a highly electronegative atom and another highly electronegative atom that lies in the neighborhood of the hydrogen atom.
For example, in water, hydrogen is covalently bonded to the oxygen atom. Therefore, hydrogen bonding arises because of the dipole-dipole interactions between the hydrogen atom of one water molecule and the oxygen atom of another water molecule.
The solubility of ether in water depends upon the extent of the formation of hydrogen bonds with water. Ether which contains three carbon atoms is soluble in water due to these lower hydrocarbon atoms can form hydrogen bonding with water.
But the solubility of hydrocarbons or ethers decreases as increase the number of carbon atoms. This is because higher ethers or ethers with more carbons have more hydrophobic parts. Therefore they cannot be soluble in water as they cannot form hydrogen bonds with water molecules.
Learn more about hydrogen bonding, here:
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Answer:
True
Hydrogen bond is a partial intermolecular bonding interaction between a lone pair on an electron rich donor atom, particularly the second-row elements nitrogen (N), oxygen (O), or fluorine (F), and the antibonding orbital of a bond between hydrogen (H) and a more
electronegative atom or group. Such an interacting system is generally denoted Dn–H···Ac, where the solid line denotes a polar covalent bond, and the dotted or dashed line indicates the hydrogen bond. The use of three centered dots for the hydrogen bond is specifically recommended by the IUPAC. While hydrogen bonding has both covalence and electrostatic contributions, and the degrees to which they contribute are currently debated, the present evidence strongly implies that the primary contribution is covelant.
Hydrogen bonds can be intermolecular (occurring between separate molecules) or
intramolecular (occurring among parts of the same molecule)
Answer:
Ag(s):H2O(l) = 3:2
For 3 moles Ag(s) we'll have 2 moles H2O(l)
Option D is correct
Explanation:
Step 1: Balancing the equation
3 Ag (s) + 4 HNO3 (aq) → 3 AgNO3 (aq) + NO (g) + 2 H2O (l)
3Ag(s) + 4NO ^3- + 4H+ →3Ag+ +3NO3- + +NO + 2H2O
3Ag(s) + NO ^3-(aq) + 4H+(aq) →3Ag+(aq) +NO(g) + 2H2O(l)
Step 2: The ratio between Ag(s) and H2O(l)
Ag(s):H2O(l) = 3:2
For 3 moles Ag(s) we'll have 2 moles H2O(l)
Option D is correct
What is she most known for?
Write down three interesting facts about her
life.
Answer:
See explanation
Explanation:
Hypatia is popular for her work in mathematics. She also did some work in the area of astronomy. Her well know work in mathematics is her ideas about conic sections.
She was born the Theon of Alexandria and she was a professional mathematician in her life time.
She was the greatest mathematician of her time and she was telling leader of the Neoplatonist school of philosophy in Alexandria. By so doing, she conquered the culture of sexism in her time.
She was trained by her father in mathematics and eventually replaced him. She was the last major mathematician in the Alexandrian tradition.
Answer:
A range of organic molecules can undergo combustion. Pyridine (C5H5 combustion in the unbalanced reaction shown below wtar o 4 CsH5N + O2 +H2O + CO2 + NO a) Write the balanced equation. (2 points) # 41 CH N +170 70 the flow, t- b) Find the percent yield for the reaction if 10.0 g of pyridine dioxide. (2 points)
Explanation:
hope this helps!
moles?
765.75
................................
The chemist uses 11 g of CaF2 and an excess of H2SO4, and the reaction produces 2.2 g of HF.
(a) Calculate the theoretical yield of HF.
(b) Calculate the percent yield of HF.
Answer:
39.3%
Explanation:
CaF2 + H2SO4 --> CaSO4 + 2HF
We must first determine the limiting reactant, the limiting reactant is the reactant that yields the least number of moles of products. The question explicitly says that H2SO4 is in excess so CaF2 is the limiting reactant hence:
For CaF2;
Number of moles reacted= mass/molar mass
Molar mass of CaF2= 78.07 g/mol
Number of moles reacted= 11g/78.07 g/mol = 0.14 moles of Calcium flouride
Since 1 mole of calcium fluoride yields two moles of 2 moles hydrogen fluoride
0.14 moles of calcium fluoride will yield 0.14×2= 0.28 moles of hydrogen fluoride
Mass of hydrogen fluoride formed (theoretical yield) = number of moles× molar mass
Molar mass of hydrogen fluoride= 20.01 g/mol
Mass of HF= 0.28 moles × 20.01 g/mol= 5.6 g ( theoretical yield of HF)
Actual yield of HF was given in the question as 2.2g
% yield of HF= actual yield/ theoretical yield ×100
%yield of HF= 2.2/5.6 ×100
% yield of HF= 39.3%