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Burka [1]
2 years ago
15

How many grams are found from 2 moles of glucose C6H1206?

Chemistry
1 answer:
Kitty [74]2 years ago
3 0

Answer:

360g

Explanation:

Hope it helps you!

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What is the amount of heat released by 1.00 gram of liquid water at 0°C when it changes to 1.00 gram of ice at 0°C?
QveST [7]

Answer:

334J/g

Explanation:

Data obtained from the question include:

Mass (m) = 1g

Specific heat of Fusion (Hf) = 334 J/g

Heat (Q) =?

Using the equation Q = m·Hf, we can obtain the heat released as follow:

Q = m·Hf

Q = 1 x 334

Q = 334J

Therefore, the amount of heat released is 334J

8 0
3 years ago
Nitrosyl bromide decomposes according to the chemical equation below. 2NOBr(g) 2NO(g) + Br2(g) 1.00 atm of NOBr is sealed in a f
Andrew [12]

Given :

2NOBr(g) - -> 2NO(g) + Br2(g)

Initial pressure of NOBr , 1 atm .

At equilibrium, the partial pressure of NOBr is 0.82 atm.

To Find :

The equilibrium constant for the reaction .

Solution :

             2NOBr(g) - -> 2NO(g) + Br2(g)

t=0 s           1 atm                 0             0

t=t_{eqb}       1( 1-2x)               2x           x

So ,

1-2x=0.82\\\\x=0.09

At equilibrium :

K_{eq}=\dfrac{[NO]^2[br_2]}{[NOBr]^2}\\\\K_{eq}=\dfrac{0.18^2\times 0.9}{0.82^2}\\\\K_{eq}=0.043\ atm

Hence , this is the required solution .

3 0
3 years ago
What is the thermal energy needed to completely melt 9.60 mol of ice at 0.0 C
inysia [295]
The  thermal energy  needed to  completely  melt  9.60  mole  of ice  at  0.0 C  is    57.8   Kj

        Explanation
ice  melt  to  form   water

The molar  heat of fusion  for   water  is  6.02  Kj/mol
Thermal energy =  moles  x  molar  heat  of  fussion  for water

=9.6  mol  x6.02 kj/mol =57.8  Kj
4 0
3 years ago
Read 2 more answers
The three isomers of pentane have different
8_murik_8 [283]
The three isomers of pentane have different structural formulas.
8 0
3 years ago
Read 2 more answers
Calculate either [ H 3 O + ] or [ OH − ] for each of the solutions.
STALIN [3.7K]

Answer: Solution A : [H_3O^+]=0.300\times 10^{-7}M

Solution B : [OH^-]=0.107\times 10^{-5}M

Solution C : [OH^-]=0.177\times 10^{-10}M

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.

pH=-\log[H_3O^+]

pOH=-log[OH^-}

pH+pOH=14

[H_3O^+][OH^-]=10^{-14}

a. Solution A: [OH^-]=3.33\times 10^{-7}M

[H_3O^+]=\frac{10^{-14}}{3.33\times 10^{-7}}=0.300\times 10^{-7}M

b. Solution B : [H_3O^+]=9.33\times 10^{-9}M

[OH^-]=\frac{10^{-14}}{9.33\times 10^{-9}}=0.107\times 10^{-5}M

c. Solution C : [H_3O^+]=5.65\times 10^{-4}M

[OH^-]=\frac{10^{-14}}{5.65\times 10^{-4}}=0.177\times 10^{-10}M

7 0
2 years ago
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