The answer is 267.93 g
Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol
The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%
Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g
Answer:
1.45 x 10⁻² g CO₂
Explanation:
To find the mass of carbon dioxide, you need to (1) convert grams CH₄ to moles CH₄ (via molar mass), then (2) convert moles CH₄ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams CO₂ (via molar mass). The final answer should have 3 sig figs to reflect the given value (5.30 x 10⁻³ g).
Molar Mass (CH₄): 12.011 g/mol + 4(1.008 g/mol)
Molar Mass (CH₄): 16.043 g/mol
Combustion of Methane:
1 CH₄ + 2 O₂ ---> 2 H₂O + 1 CO₂
Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)
Molar Mass (CO₂): 44.007 g/mol
5.30 x 10⁻³ g CH₄ 1 mole 1 mole CO₂ 44.007 g
--------------------------- x ---------------- x --------------------- x ----------------- =
16.043 g 1 mole CH₄ 1 mole
= 0.0145 g CO₂
= 1.45 x 10⁻² g CO₂
Answer:
The answer is "Greater than zero, and greater than the rate of the reverse reaction".
Explanation:
It applies a rate of reaction to the balance, a forward response dominates until it reaches a constant. This process is balanced before 52 mmol of the reactant
, to which 3 is added. In balance, that rate of the forward reaction was its rate with forwarding reaction, both of which are higher than 0 as the response has achieved balance so that both species get a level greater than 0.
Answer: The correct answer is "B" two bonding domains(or bonding pairs) or two non bonding domains(or lone pairs)
Explanation:
Molecular geometry/structure is a three dimensional shape of a molecule. It is basically an arrangement of atoms in a molecule.It is determined by the central atom, its surrounding atoms and electron pairs.According to VSEPR theory, there are 5 basic shapes of a molecule: linear, trigonal planar, tetrahedral, trigonal bipyramidal and octahedral.
A)Four bonding domains and zero non bonding domains: shape is tetrahedral and bond angle is 109.5°
B)Two bonding domains and two non bonding domains(lone pairs): shape is bent and bond angle is 104.5°
C)Three bonding domains and one non bonding domain: shape is trigonal pyramidal and bond angle is 107°
D)Two bonding domain and zero non bonding domain: shape is linear and bond angle is 107°
E)Two bonding domain and one non bonding domain: bent shape and bond angle is 120°
F)Three bonding domains and zero nonbonding domain: shape is trigonal planar and bond angle is 120°
Hence Two bonding domains and two non bonding domains have the smallest bond angle.
