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Semenov [28]
2 years ago
6

How many moles are in 9.12 x 1023 molecules of sugar?​

Chemistry
1 answer:
aleksandrvk [35]2 years ago
3 0

Answer:

1.514 moles

Explanation:

For this problem you want to use dimensional analysis and cancel out your molecules of sugar and be left with moles of sugar. We know that 1 mole (of anything) = 6.022 x 10 ^ 23 molecules, so we should use that conversion to help us. Start with 9.12 x 10 ^23 molecules and divide by 6.022 x 10 ^ 23 molecules, and you will be left with moles.

Hope this helps!

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The question is below
Nezavi [6.7K]
The correct conditions for measuring reduction potentials (the tendency to acquire electrons and become reduced), is 25C and 1M (or 1 mole/litre) for reactants - ANSWER B.
3 0
3 years ago
2. Which types of changes observe the law of
mart [117]
D) physical and chemical changes
3 0
3 years ago
How many moles of Manganese there in are 5.76 x 10(15) atoms of Mn?
aksik [14]

Answer:

1. 9.57 × 10^-9 moles.

2. 7.38mol

Explanation:

1.) To find the number of moles there are in the number of particles in an atom, we divide the number of particles (nA) by Avagadro's constant (6.02 × 10^23)

Hence, to find the number of moles (n) of Manganese (Mn), we say:

5.76 x 10^15 atoms ÷ 6.02 × 10^23

5.76/6.02 × 10^(15-23)

= 0.957 × 10^-8

= 9.57 × 10^-9 moles.

2.) Mole = mass/molar mass

Molar mass of sodium chloride (NaCl) = 23 + 35.5

= 58.5g/mol

mole = 431.6 g ÷ 58.5g/mol

mole = 7.38mol

7 0
3 years ago
Please help I need the awnser quick will give brain to first to awnser
gizmo_the_mogwai [7]
Possibly decomposition but not sure
7 0
3 years ago
Read 2 more answers
At 150°C the decomposition of acetaldehyde CH3CHO to methane is a first order reaction. If the
Crank

The decomposition time : 7.69 min ≈ 7.7 min

<h3>Further explanation</h3>

Given

rate constant : 0.029/min

a concentration of  0.050 mol L  to a concentration of 0.040 mol L

Required

the decomposition time

Solution

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time

For first-order reaction :

[A]=[Ao]e^(-kt)

or

ln[A]=-kt+ln(A0)

Input the value :

ln(0.040)=-(0.029)t+ln(0.050)

-3.219 = -0.029t -2.996

-0.223 =-0.029t

t=7.69 minutes

4 0
3 years ago
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