Since the temperature
is a constant, we can use Boyle's law to solve this.<span>
<span>Boyle' law says "at a constant temperature, the
pressure of a fixed amount of an ideal gas is inversely proportional to its
volume.
P α 1/V
</span>⇒
PV = k (constant)<span>
Where, P is the pressure of the gas and V is the
volume.
<span>Here, we assume that the </span>gas in the balloon is an ideal gas.
We can use Boyle's law for these two situations as,
P</span>₁V₁ = P₂V₂<span>
P₁ = 100.0 kPa = 1 x 10⁵ Pa
V₁ =
3.3 L
P₂ =
90.0 x 10³ Pa
V₂ =?
By substitution,
1 x 10⁵ Pa x 3.3 L = 90 x 10³ Pa x V₂</span><span>
V</span>₂ = 3.7 L<span>
</span><span>Hence, the volume of gas when pressure is 90.0 kPa
is 3.7 L.</span></span>
Answer:
I believe that it would be copper because copper is all of those things.
Explanation:
Answer:
Explanation:
Potassium nitrate is a soluble salt which readily dissolves in a polar solvent, such as water. When solid potassium nitrate is dissolved in water, it dissociates into potassium cations and nitrate anions.
Due to the resultant ionic charges, the polar water molecules attract the resultant ions and potassium nitrate ions become hydrated, that is, surrounded by water molecules.
Nitrate, our anion, attracts the partially positive ends of water molecules by attracting them via hydrogen atom.
Potassium, the cation, attracts the partially negative end of water molecules by attracting via oxygen atom.
Answer:
Density is a measurement that compares the amount of matter an object has to its volume. An object with much matter in a certain volume has high density.
Explanation:
Answer:
We need 1.1 grams of Mg
Explanation:
Step 1: Data given
Volume of water = 78 mL
Initial temperature = 29 °C
Final temperature = 78 °C
The standard heats of formation
−285.8 kJ/mol H2O(l)
−924.54 kJ/mol Mg(OH)2(s)
Step 2: The equation
The heat is produced by the following reaction:
Mg(s)+2H2O(l)→Mg(OH)2(s)+H2(g)
Step 3: Calculate the mass of Mg needed
Using the standard heats of formation:
−285.8 kJ/mol H2O(l)
−924.54 kJ/mol Mg(OH)2(s)
Mg(s) + 2 H2O(l) → Mg(OH)2(s) + H2(g)
−924.54 kJ − (2 * −285.8 kJ) = −352.94 kJ/mol Mg
(4.184 J/g·°C) * (78 g) * (78 - 29)°C = 15991.248 J required
(15991.248 J) / (352940 J/mol Mg) * (24.3 g Mg/mol) = 1.1 g Mg
We need 1.1 grams of Mg
