In seawater, salt is the solute and water is the solvent.
Answer:

Explanation:
1. Volume of sealed tube
Assume the sealed tube is a right circular cylinder in which the cap and the base are also 4.20 mm thick.
Its outside dimensions are 155 mm long × 10.0 mm diameter.
Its inside dimensions are
h = 155 mm - 2 × 4.20 mm = 146.6 mm
r = 5.0 mm - 4.20 mm = 0.8 mm
V = πr²h = π(0.8)²× 146.6 mm³ = 294.8 mm³ = 0.2948 cm³
2. Calculate the mass of NH₃
Volume of H2 produced = 57.6576 L
<h3>Further explanation</h3>
Given
23.17 g Be
Required
Volume of H2
Solution
Reaction
Be(s)+H2O(g)→BeO(s)+H2(g)
mol Be :
= 23.17 g : 9 g/mol
= 2.574
From the equation, mol H2 : mol Be = 1 : 1, so mol H2 = 2.574
Volume H2(assumed at STP, 1 mol=22.4 L) :
= 2.574 x 22.4 L
= 57.6576 L
Answer:
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Explanation: