Answer:
247 ml
Explanation:
How many mL of 0.150 M HF solution are required to produce 0.0370 moles of HF 0.150 moles/ liter = 0.150/1000 moles/ml =0.000150 moles/ml
0.000150 x ? = 0.0370 moles HF
? = 0.0370/0.000150 = 247 ml
check
247 ml = 247/1000 L = 0.247
(0.247) x (0.150) =0.370 check
Answer:
Explanation:
Polyatomic ions are ions (usually anions) that are made up of more than one atom. In order to determine the charge of anions, one can first identify the charge of the metal (which forms the cation) in a compound (which is usually easy to know) before predicting the charge of the anion. The charges are usually exchanged to form denominator of the other reacting atom/molecule, but if divisible, they are divided first before the exchange.
(a) The polyatomic ion in KNO₂ is NO₂ with the ionic equation below showing it's charge
KNO₂ ⇒ K⁺ + NO₂⁻
From the above, we can deduce that the charge of NO₂ is "1-"
(b) The polyatomic ion in CaSO₄ is SO₄ with the ionic equation below showing it's charge
CaSO₄ ⇒ Ca²⁺ + SO₄²⁻
The charge of SO₄ from the above equation is "2-"
(c) The polyatomic ion in Mg(NO₃)₂ is NO₃ with the ionic equation below showing it's charge
Mg(NO₃)₂ ⇒ Mg²⁺ + NO₃⁻
From the above equation, it can be deduced that the charge of NO₃ is "1-"
Explanation:
Caesium is a highly electropositive metal found in the first group on the periodic table.
Electropositivity is a measure of the readiness of an atom to lose electrons. It is related to the ionization energy and electronegativity of an element.
- The ionization energy is the energy required to remove the most loosely bond electron in an atom.
- The lower the ionization energy the lower the electronegativity.
- The more electropositive or metallic it will be.
- Metals found on the left ofthe periodic table are the most reactive metal in the lower left corner. This where caesium is found.
Learn more:
Electronegativity brainly.com/question/7007192
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Answer:
[a]. - 445.45J,
[b]. - 367.92 J
Explanation:
The following parameters are given in the question above. These information are used in solving this problem.
The mass of Xenon = 65.0 g of xenon, pressure = 2.00 atm, temperature = 298 K.
The number of moles of xenon = mass/ molar mass = 65g/ 131.293= 0.495.
The cp= 3/2 R, cp =3/2R + R = 5/2 R.
j = cp/cv = 3/2.
[a]. The final temperature,T2 = (2)^-2/3 × (298)^5/3 = T2^5/3.
Final temperature,T2 = 225.84K.
Expansion work = nCv [ T₂ - T₁] = 0.495 × 3/2 × 8.314 × [ 225.84 - 298] = - 445.45J.
(b). The final temperature can be Determine as;
3/2( T2 - 298k) = - 1 (T2 /1 - 298/2).
3/2(T2 - 298) = - T2 + 149K.
3T2 - 894 = - 2T2 +298K.
T2 = 238.4 K.
Workdone= nCv (T2 - T1) = 0.495 × 3/2 × 8.314 (238.4 - 298) = - 367.92 J.