1. Dry Ice (solid carbon dioxide)
2. Iodine
3. Arsenic
4. Naphthalene
Using the answer from the first part, we know that 2.957 moles of bismuth have formed. Moreover, the molar ratio between bismuth and carbon monoxide is:
2 : 3
Using the method of ratios,
2 : 3
2.957 : CO
CO = (3 * 2.957) / 2
CO = 4.4355
4.436 moles of carbon monoxide will be formed
QPOE Files
The x-ray data are stored in QPOE files (Quick Position-Ordered Events, *.qp) rather than image arrays. These are lists of photons identified by several quantities, including the position on the detector, pulse height, and arrival time. Note that, unlike IRAF images, QPOE files have no associated header file, and are always stored in the current directory, unless explicitly specified otherwise. Non-PROS IRAF tasks can also access QPOE data files in place of image arrays.
Answer:
(a) Pair 1: H₂S and HS⁻
Pair 2: NH₃ and NH₄⁺
(b) Pair 1: HSO₄⁻ and SO₄⁻
Pair 2: NH₃ and NH₄⁺
(c) Pair 1: HBr and Br⁻
Pair 2: CH₃O⁻ and CH₃OH
(d) Pair 1: HNO₃ and NO₃⁻
Pair 2: H₃O⁺
Explanation:
When an acid loses its proton (H⁺), a conjugate base is produced.
When a base accepts a proton (H⁺), it forms a conjugate acid.
(a) H₂S is an acid. When it loses a proton, it forms the conjugate base HS⁻.
NH₃ is a base. When NH₃ gains a proton, it forms the conjugate acid NH₄⁺
(b) The acid HSO₄⁻ loses a H⁺ ion and forms the conjugate base SO₄²⁻.
The base NH₃ accepts a H⁺ ion to form the conjugate acid NH₄⁺.
(c) HBr is an acid. When loses the H⁺ ion, it forms the conjugate base Br⁻.
CH₃O⁻ accepts a H⁺ ion to form the conjugate acid CH₃OH.
(d) HNO₃ loses a proton to form the conjugate base NO₃⁻.
H₂O gains a proton to form the conjugate acid H₃O⁺.
Answer:
1) 1,... 2
2) 18
3) n= 3 and I=1
Explanation:
1) when l= 0, its an s-sub-level, and only 1 orbital is possible which can carry only 2-electrons
2) the maximum number of electron is given by 2n^2= 2×3^2= 18
3) in 3p, the coefficient of p is the value of n= 3 and l-value of P is 1
