Electronegativity of an element decreases as we move down a group on the periodic table and electronegativity increases while moving from left to right across a period on the periodic table.
Explanation:
- The electronegativity increases as we move from left to right across a period because from left to right across a period, the nuclear charge is increasing Hence the attraction for the valence electrons also increases.
- As we move down a group, the atoms of each element have an increasing number of energy levels. The distance between the nucleus and valence electron shell increases and reduces the attraction for valence electrons. Hence electronegativity decreases as we move from top to bottom down a group.
Answer:
- <u>No, you cannot dissolve 4.6 moles of copper sulfate, CuSO₄, in 1750mL of water.</u>
Explanation:
This question is part of a Post-Lab exercise sheet.
Such sheet include the saturation concentrations for several salts.
The saturation concentration of Copper Sulfate, CuSO₄, indicated in the table is 1.380M.
That means that 1.380 moles of copper sulfate is the maximum amount that can be dissolved in one liter of solution.
Find the molar concentration for 4.6 moles of copper sulfate in 1,750 mL of water.
You need to assume that the volume of water (1750mL) is the volume of the solution. This is, that the 4.6 moles of copper sulfate have a negligible volume.
<u>1. Volume in liters:</u>
- V = 1,750 mL × 1 liter / 1,000 mL = 1.75 liter
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<u>2. Molar concentration, molarity, M:</u>
- M = number of moles of solute / volume of solution in liters
- M = 4.6 moles / 1.75 liter = 2.6 M
Since the solution is saturated at 1.380M, you cannot reach the 2.6M concentration, meaning that you cannot dissolve 4.6 moles of copper sulfate, CuSO₄ in 1750mL of water.
Note: The question is incomplete. The complete question is given below :
Suppose a substance has a heat of fusion equal to 45 cal/g and a specific heat of 0.75 cal/g°C in the liquid state. If 5.0 kcal of heat are applied to a 50 g sample of the substance at a temperature of 24°C, what will its new temperate be? What state will the sample be in? (melting point of the substance = 27°C; specific heat of the solid =0.48 cal/g°C; boiling point of the substance = 700°C)
Explanation:
1.a) Heat energy required to raise the temperature of the substance to its melting point, H = mcΔT
Mass of solid sample = 50 g; specific heat of solid = 0.75 cal/g; ΔT = 27 - 24 = 3 °C
H = 50 × 0.75 × 3 = 112.5 calories
b) Heat energy required to convert the solid to liquid at its melting point at 27°C, H = m×l, where l = 45 cal/g
H = 50 × 45 = 2250 cal
c) Total energy used so far = 112.5 cal + 2250 cal = 2362.5 calories.
Amount of energy left = 5000 - 2362.5 = 2637.5 cal
The remaining energy is used to heat the liquid
H = mcΔT
Where specific heat of the liquid, c = 0.75 cal/g/°C, H = 2637.5 cal, ΔT = temperature change
2637.5 = 50 × 0.75 x ΔT
ΔT = 2637.5 / ( 50*0.75)
ΔT = 70.3 °C
Final temperature of sample = (70.3 + 27) °C = 97.3 °C
The substance will be in liquid state at a temperature of 97.3 °C
i hope that this eg gonna help u
Answer:
1. Based on the given question, 138.03 grams of NO₂ is reacting completely with 18.02 grams of H2O. However, in case when 359 grams of NO₂ is used then the grams of water consumed in the reaction will be,
= 359 × 18.02 / 138.03 = 46.87 grams of water.
2. As mentioned in the given case 138.03 grams of NO₂ generates 126.04 grams of HNO₃. Therefore, 359 grams of NO₂ will produce,
= 359 × 126.04 / 138.03
= 327.81 grams of HNO₃.
3. Based on the given question, 138.04 grams of NO₂ is generating 30.01 grams of NO. Therefore, 359 grams of NO₂ will generate,
= 359 × 30.01 / 138.04
= 78.04 grams of NO.
