The color emitted be larger atoms is lower in energy then the light emitted by smaller atoms
Answer:
31395 J
Explanation:
Given data:
mass of water = 150 g
Initial temperature = 25 °C
Final temperature = 75 °C
Energy absorbed = ?
Solution:
Formula:
q = m . c . ΔT
we know that specific heat of water is 4.186 J/g.°C
ΔT = final temperature - initial temperature
ΔT = 75 °C - 25 °C
ΔT = 50 °C
now we will put the values in formula
q = m . c . ΔT
q = 150 g × 4.186 J/g.°C × 50 °C
q = 31395 J
so, 150 g of water need to absorb 31395 J of energy to raise the temperature from 25°C to 75 °C .
The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL
<h3>Balanced equation </h3>
2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O
From the balanced equation above,
- The mole ratio of the acid, H₃PO₄ (nA) = 2
- The mole ratio of the base, Ca(OH)₂ (nB) = 3
<h3>How to determine the volume of Ca(OH)₂ </h3>
- Molarity of acid, H₃PO₄ (Ma) = 0.390 M
- Volume of acid, H₃PO₄ (Va) = 24.5 mL
- Molarity of base, Ca(OH)₂ (Mb) = 0.279 M
- Volume of base, Ca(OH)₂ (Vb) =?
MaVa / MbVb = nA / nB
(0.39 × 24.5) / (0.279 × Vb) = 2/3
9.555 / (0.279 × Vb) = 2/3
Cross multiply
2 × 0.279 × Vb = 9.555 × 3
0.558 × Vb = 28.665
Divide both side by 0.558
Vb = 28.665 / 0.558
Vb = 51.4 mL
Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL
Learn more about titration:
brainly.com/question/14356286
Answer:
Ammonium bromide can be prepared by the direct action of hydrogen bromide on ammonia. It can also be prepared by the reaction of ammonia with iron(II) bromide or iron(III) bromide, which may be obtained by passing aqueous bromine solution over iron filings.
Explanation:
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Answer:
0.50 M
Explanation:
Given data
- Mass of sodium sulfate (solute): 7.1 g
- Volume of solution: 100 mL
Step 1: Calculate the moles of the solute
The molar mass of sodium sulfate is 142.04 g/mol. The moles corresponding to 7.1 grams of sodium sulfate are:
Step 2: Convert the volume of solution to liters
We will use the relation 1 L = 1000 mL.
Step 3: Calculate the molarity of the solution
