Answer: 23.8889
Explanation:
(75°F − 32) × 5/9 = 23.889°C
Answer:
Nickel (Ni) is most likely to form more than one kind of positively charged ion as it is a transition metal and all of the ther metals have a fixed ion charge.
Na becomes Na+
Ba becomes Ba +2
Mg becomes Mg+2
Explanation:
please mark brailiest
So the first thing we must do is write a balanced equation for the reaction and we know the equation is balnced when all the species on the RHS is equal to the species on the LHS
2NaOH + H₂SO₄ → Na₂SO₄<span>
+ 2H₂O</span>
So now it's time to identify what reactant you know the most for from the question (volume & conc. of H₂SO₄) and use that info to find the unknown (conc. of NaOH)
If 1000 ml of H₂SO₄ contain 0.750 mol [0.750 M is the amount of moles in
1 L (1000 ml)]
then let 15 ml of H₂SO₄ contain x mol [15 ml is the amount of the acid that took part in the reaction]
⇒
x =
= 0.01125 molMole ratio of NaOH to H₂SO₄ can be obtained from the balanced equation
0
2NaOH +
1H₂SO₄ → Na₂SO₄ + 2H₂O
mole ratio of NaOH to H₂SO₄ is 2 : 1∴ if mole of of H₂SO₄ = 0.01125 mol then moles of NaOH = (0.01125 mol) × 2 = 0.0225 molIf 17.5 ml of NaOH contain 0.0225 mol [this was given in the question]
then let 1000 ml of NaOH contain x⇒ x =
= 1.286 mol∴ concentration of NaOH is 1.286 mol/L
Explanation:
Pressure of a gas is the combined force with which the molecules bombard a unit area of the wall of the container.
1 atm = 760mmHg
= 760torr
= 101325Pa
= 101325Nm⁻²
1mmHg = 1torr
1Pa = 1Nm⁻²
101.325 kPa and 101,325 Pa are the same
1000Pa = 1kPa
101,325 Pa and 1 atm
1atm and 101.325 kPa
4.10 g C12H22O11 * ( 1 mol / 342.2965 g ) * ( 6.022x10^23 / 1 mol ) = <span>7.2107077x10^21 molecules = 7.21x10^21 molecules
Convert grams to moles using the molar mass ( g/mol ).
Use Avogadro's constant ( 6.022x10^23 ) to get the quantity of molecules from the amount of moles.
Revise answer to have the correct number of significant figures ( the smallest amount of digits in any number in the problem is the amount of significant figures your answer should have ).</span>