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I am Lyosha [343]
3 years ago
6

Check all that apply to sugar.

Chemistry
2 answers:
Sav [38]3 years ago
6 0

Answer:

Does not conduct electricity and is a non-electrolyte

Explanation:

Because I pay attention

harina [27]3 years ago
3 0

Answer:

Candy

Explanation:

its sweet and has a lot of sugar and acid

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If an atom contains 12 protons and 10 neutrons, what's its mass number?
user100 [1]
A = Z + n

A = 12 + 10

A = 22

Answer C

hope this helps!
7 0
3 years ago
Calculate the volume of an object that has a density of 4 g/mL and has a mass of 128 grams. Show your work
SOVA2 [1]
The first one is 32mL and the second one is 2.62 and I think it’s grams/mL I’m not for sure about the letters on the second one

4 0
2 years ago
The reaction that occurs in a Breathalyzer, a device used to determine the alcohol level in a person's bloodstream, is given bel
Umnica [9.8K]

Answer:

The rate of disappearance of C₂H₆O = 2.46 mol/min

Explanation:

The equation of the reaction is given below:

2 K₂Cr₂O₇ + 8 H₂SO₄ + 3 C₂H₆O → 2 Cr₂(SO₄)₃ + 2 K₂SO₄ + 11 H₂O

From the equation of the reaction, 3 moles of C₂H₆O is used when 2 moles of Cr₂(SO₄)₃ are produced, therefore, the mole ratio of C₂H₆O to Cr₂(SO₄)₃ is 3:2.

The rate of appearance of Cr₂(SO₄)₃ in that particular moment is given 1.64 mol/min. This would than means that C₂H₆O must be used up at a rate which is approximately equal to their mole ratios. Thus,  the rate of of the disappearance of C₂H₆O can be calculated from the mole ratio of Cr₂(SO₄)₃ and C₂H₆O.

Rate of disappearance of C₂H₆O = 1.64 mol/min of  Cr₂(SO₄)₃  * 3 moles of C₂H₆O / 2 moles of Cr₂(SO₄)₃

Rate of disappearance of C₂H₆O = 2.46 mol/min of C₂H₆O

Therefore, the rate of disappearance of C₂H₆O = 2.46 mol/min

4 0
3 years ago
For the reaction C2H4(g) + H2O(g) --> CH3CH2OH(g)
Dominik [7]

Answer : The value of equilibrium constant for this reaction at 262.0 K is 3.35\times 10^{2}

Explanation :

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

where,

\Delta G^o = standard Gibbs free energy  = ?

\Delta H^o = standard enthalpy = -45.6 kJ = -45600 J

\Delta S^o = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

Now put all the given values in the above formula, we get:

\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)

\Delta G^o=-12666.6J=-12.7kJ

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln k

where,

\Delta G^o = standard Gibbs free energy  = -12666.6 J

R = gas constant  = 8.314 J/K.mol

T = temperature  = 262.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k

k=3.35\times 10^{2}

Therefore, the value of equilibrium constant for this reaction at 262.0 K is 3.35\times 10^{2}

3 0
3 years ago
Identify the limiting reactant when 32. 0 g hydrogen is allowed to react with 16. 0 g oxygen
Mkey [24]

Answer:

Oxygen is limiting reactant

Explanation:

2 H2  +   O2  ======> 2 H2 O

from this equation (and periodic table) you can see that

  4 gm of H combine with 32 gm O2  

     H / O  =  4/32 = 1/8

       32 /16    =  2/1    shows O is limiter

         for 32 gm H you will need 256 gm O   and you only have 16 gm

3 0
1 year ago
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