Answer:
energy required is 0.247kJ
Explanation:
The formula to use is Energy = nRdT;
Where n is number of mole
R is the molar gas constant
dT is the change in temperature
n = reacting mass of mercury / molar mass of mercury = 27.4/200.59 = 0.137
dT = final temperature - initial temperature = 376.20 - 158.30 = 217.90K
R = 8.314Jper mol per Kelvin
Energy = 0.137 x 8.314 x 217.90 = 247.12J
Energy in kJ= 247.12/1000= 0.247kJ
Answer: gas particles that are in constant motion and exhibit perfectly elastic collisions
Explanation:
The correct term to fill in the blank would be decrease. Phytic acid, tannins, and oxalates decrease the absorption of many minerals. Phytic acid reduce the absorption of a number of minerals like iron, zinc and calcium. Also, it would encourage mineral deficiencies. The same is also true for tannins and oxalates.
a.
Acids react with bases and give salt and water and the products.
Hence, HCl reacts with NaOH and gives NaCl salt and H₂O as the products. The reaction is,
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
To balance the reaction equation, both sides hould have same number of elements.
Left hand side, Right hand side,
H atoms = 2 H atoms = 2
Cl atoms = 1 Cl atoms = 1
Na atoms = 1 Na atoms = 1
O atoms = 1 O atoms = 1
Hence, the reaction equation is already balanced.
b.
Molarity (M)= moles of solute (mol) / Volume of the solution (L)
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
Molarity of NaOH = <span>0.13 M
</span>Volume of NaOH added = <span>43.7 mL
Hence, moles of NaOH added = 0.13 M x 43.7 x 10</span>⁻³ L
= 5.681 x 10⁻³ mol
Stoichiometric ratio between NaOH and HCl is 1 : 1
Hence, moles of HCl = moles of NaOH
= 5.681 x 10⁻³ mol
5.681 x 10⁻³ mol of HCl was in <span>26.9 mL.
Hence, molarity of HCl = </span>5.681 x 10⁻³ mol / 26.9 x 10⁻³ L
= 0.21 M
Answer: A volume of 58.69 mL of water would you add to 35.81 mL of 1.59 M acetic acid solution to make it a 0.97 M.
Explanation:
Given: 
= 35.81 mL, 
= 1.59 M
= ?, 
= 0.97 M
Formula used to calculate volume is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that a volume of 58.69 mL of water would you add to 35.81 mL of 1.59 M acetic acid solution to make it a 0.97 M.
