(4 mol H2O) x (112 kJ / 3 mol H2O) = 149 kJ
<span>(14.5 g HCl) / (36.4611 g HCl/mol) x (112 kJ / 3 mol HCl) = 14.9 kJ </span>
<span>(475 kJ) / (181 kJ / 2 mol HgO) x (216.5894 g HgO/mol) = 1137 g HgO </span>
<span>(179 kJ) / (181 kJ / 1 mol O2) x (31.99886 g O2/mol) = 31.6 g O2 </span>
<span>(145 kJ) / (112 kJ / 3 mol Cl2) x (70.9064 g Cl2/mol) = 275 g Cl2 </span>
<span>(14.5 g S2Cl2) / (135.0360 g S2Cl2/mol) x (112 kJ / 1 mol S2Cl2) = 12.0 kJ </span>
<span>CaCO3 + 2 NH3 → CaCN2 + 3 H2O; ∆H = –90.0 kJ </span>
<span>(798 kJ) / (90.0 kJ / 2 mol HN3) x (17.03056 g NH3/mol) = 302 g NH3 </span>
<span>(19.7 g H2O) / (18.01532 g H2O/mol) x (90.0 kJ / 3 mol H2O) = 32.8 kJ</span>
Answer:
37.98 kPa.
Explanation:
- We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and T are constant, and have different values of P and V:
<em>(P₁V₁) = (P₂V₂)</em>
<em></em>
P₁ = 101.3 kPa, V₁ = 1.5 L,
P₂ = ??? kPa, V₂ = 4.0 L.
- Applying in the above equation
<em>(P₁V₁) = (P₂V₂)</em>
<em></em>
<em>∴ P₂ = (P₁V₁)/V₂</em> = (101.3 kPa)(1.5 L)/(4.0 L) = <em>37.98 kPa.</em>
Answer:
La réponse ce trouve en bas
Explanation:
13 a) Oui, ce possible parce la solubilité du sel dans l'eau a 60° est majeur que 50g du sel en 150 ml d'eau (333.3 g/L). Dans la graphique la solubilité est de 370 g/L.
b) Jusqu'au 30°C, (30°, 360g(L), ce trouvent ces points dans la ligne rouge.
Answer:
a. The number of neutrons in the nucleus
Explanation:
Having more protons or electrons would change the charge of the atom making it an ion. Neutrons have no charge so the overall charge of the atom is not affected; neutrons also have a relative mass of 1 which is why there are different isotopes.
Hope this helps!
Answer:
The answer to be filled in the respective blanks in question is
3 and 1
Explanation:
So, we know that the formation of cabon-dioxide mole and that of Adenosin-Tri-Phosphate (ATP) moles will be in the ratio of 3:1 i.e., three carbon-di-oxide moles and 1 ATP mole.
Therefore, we can say that one pyruvate mole when passed through citric acid cycle and pyruvate dehydrogenase yields carbon-di-oxide and ATP moles in the ratio 3:1
