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kari74 [83]
2 years ago
12

What is Oxidation state of sulphur in Na2S4O6?Help me..​

Chemistry
1 answer:
quester [9]2 years ago
6 0

Answer:

oxidation state of sulphur=x

Explanation:

Na2S4O6=2[+1]+4x+6[-2]=0

                 +2+4x-12=0

                     4x-10=0

                           4x=10

                            x=10/4=2.5

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the chemical formula for copper iodide is Cul. How many atoms of copper(Cu) is it made from? How many atoms of iodine(I)?
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2 years ago
When the cell splits, they have a nucleus with complete genetic material for the new daughter cells. What is this genetic materi
vaieri [72.5K]
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3 years ago
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7 0
3 years ago
Calculate the molarity of a solution of Nach if it contains 7.2.g Nach in 100.0 mL of solution. andver: m Nach . .
Norma-Jean [14]

Answer:

1.23 M

Explanation:

Molarity of a substance , is the number of moles present in a liter of solution .

M = n / V

M = molarity  

V = volume of solution in liter ,

n = moles of solute ,

Moles is denoted by given mass divided by the molecular mass ,  

Hence ,  

n = w / m

n = moles ,  

w = given mass ,  

m = molecular mass .

From the question ,

w = given mass of NaCl = 7.2 g

As we know , the molecular mass of NaCl = 58.5 g/mol

Moles is calculated as -

n = w / m  =  7.2 g / 58.5 g/mol = 0.123 mol

Molarity is calculated as -

V = 100ml = 0.1 L            (since , 1 ml = 1/1000L )

M = n / V  =  0.123 mol / 0.1 L = 1.23 M

5 0
3 years ago
The enzyme, phosphoglucomutase, catalyzes the interconversion
Fittoniya [83]

Answer:

K_{eq = 19

ΔG° of the reaction forming glucose 6-phosphate =  -7295.06 J

ΔG° of the reaction  under cellular conditions = 10817.46 J

Explanation:

Glucose 1-phosphate     ⇄     Glucose 6-phosphate

Given that: at equilibrium, 95% glucose 6-phospate is  present, that implies that we 5% for glucose 1-phosphate

So, the equilibrium constant K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq= \frac{0.95}{0.05}

K_{eq = 19

The formula for calculating ΔG° is shown below as:

ΔG° = - RTinK

ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k ×  1n(19))

ΔG° = 7295.05957 J

ΔG°≅ - 7295.06 J

b)

Given that; the concentration  for  glucose 1-phosphate = 1.090 x 10⁻² M

the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M

Equilibrium constant  K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq}= \frac{1.395*10^{-4}}{1.090*10^{-2}}

K_{eq} = 0.01279816514  M

K_{eq} = 0.0127 M

ΔG° = - RTinK

ΔG° = -(8.314*298*In(0.0127)

ΔG° = 10817.45913 J

ΔG° = 10817.46 J

5 0
2 years ago
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