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ololo11 [35]
3 years ago
15

To name the compound written as cobr2, you would write:

Chemistry
2 answers:
inna [77]3 years ago
8 0

cobalt(II) bromide. (apex)

Tcecarenko [31]3 years ago
7 0
Answer is: cobalt(II) bromide.

Bromine has oxidation number -1 (because it is nonmetal from goup XVII and gain one electron to have electron configuration as closestgas noble) , so cobalt must have oxidation number +2, because compound has neutral charge: 
x + 2·(-1) = 0.
x = +2.
Cobalt(II) bromide<span> red, solid</span><span> inorganic compound, soluble in water.</span>

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To help them discover new constructive forces on earth
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______ is the use of enzymes is necessary to increase the activation energy requirements of a chemical reaction.
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Physostigmine, an alkaloid obtained from a West African plant (Physotigma venenosum),is used in the treatment of glaucoma. Treat
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The structure of the salt is attached.

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Physostigmine, an alkaloid obtained from a West African plant (Physotigma venenosum),is used in the treatment of glaucoma. Treatment of physostigmine with methyl iodide gives a quaternary ammonium salt. What is the structure of this salt?

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...because this notation increases the convenience in using the numbers 
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Read 2 more answers
A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a
Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

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