Answer:
b. 485 kPa
Explanation:
Gay-Lussac's law express that the pressure of a gas under constant volume is directly proportional to the absolute temperature. The equation is:
P1T2 = P2T1
<em>P is pressure and T absolute temperature of 1, initial state and 2, final state of the gas</em>
<em>Where P1 = 74psi</em>
<em>T2 = 20°C + 273.15 = 293.15K</em>
<em>P2 = ?</em>
<em>T1 = (95°F -32) * 5/9 + 273.15 = 308.15K</em>
<em />
Replacing:
74psi*293.15K = P2*308.15K
70.4psi
In kPa:
70.4psi * (6.895kPa / 1psi) =
<h3>b. 485 kPa
</h3>
Answer:
2.down is hypotheses and 23.across is experiment
Explanation:
akechis pancakes
A grounding electrode is any object that directly links to the earth. They are most times used to divert electricity from the elements.
- Swimming pool structures and structural <u>reinforcing steel. 250.52(B)(3)</u><u>,</u> [680.26(B)(1), and (B)(2)] shall not be used as a grounding electrode.
In code 250.52(B)(3) it is clearly specified that the bonding grid and reinforcing steel that is related to a pool should not be used as grounding electrodes.
This is essential because when a metal that lies beneath a swimming pool is used as a grounding electrode, current from nearby electrical systems can be introduced into the pool.
This could cause the electrocution of anybody in the swimming pool at that time.
Learn more here:
brainly.com/question/14681208
The balanced equation between NaOH and H₂SO₄ is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of moles of NaOH moles reacted = molarity of NaOH x volume
number of NaOH moles = 0.08964 mol/L x 27.86 x 10⁻³ L = 2.497 x 10⁻³ mol
according to molar ratio of 2:1
2 mol of NaOH reacts with 1 mol of H₂SO₄
therefore 2.497 x 10⁻³ mol of NaOH reacts with - 1/2 x 2.497 x 10⁻³ mol of H₂SO₄
number of moles of H₂SO₄ reacted - 1.249 x 10⁻³ mol
Number of H₂SO₄ moles in 34.53 mL - 1.249 x 10⁻³ mol
number of H₂SO₄ moles in 1000 mL - 1.249 x 10⁻³ mol / 34.53 x 10⁻³ L = 0.03617 mol
molarity of H₂SO₄ is 0.03617 M
Yes, it is a special case of enthalpy of neutralization.
The enthalpy of neutralization (ΔHn) is the change in enthalpy that occurs when one equivalent of an acid and one equivalent of a base undergo a neutralization reaction to form water and a salt.
The standard enthalpy change of neutralization is the enthalpy change when solutions of an acid and an alkali react together under standard conditions to produce 1 mole of water.
