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Sloan [31]
3 years ago
8

Hope is testing a substance using blue litmus paper. The paper turns red. The substance is _____.

Chemistry
2 answers:
garri49 [273]3 years ago
6 0
An acid that’s the answer!
Aleksandr [31]3 years ago
3 0

Answer:

the substance is an acid

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An alkane with the formula C6H14 can be prepared by hydrogenation of either of two precursor alkenes having the formula C6H12.
ankoles [38]
Two precursor alkenes

    H₃C  CH₃
         I   I
H₂C=C-CH-CH₃    2,3-dimethyl-1-butene

   H₃C      CH₃
        I       I
H₃C-CH=CH-CH₃    2,3-dimethyl-2-butene


alkane

   H₃C     CH₃
        I      I
H₃C-CH-CH-CH₃    2,3-dimethylbutane

    H₃C  CH₃                        H₃C     CH₃
         I   I                                   I      I
H₂C=C-CH-CH₃  + H₂ → H₃C-CH-CH-CH₃ 
  
    H₃C   CH₃                       H₃C     CH₃
        I    I                                   I      I
H₂C-C=CH-CH₃  + H₂ → H₃C-CH-CH-CH₃ 

6 0
3 years ago
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The student wants to test the conductivity of each solution. Prior to carrying out the investigation, the student needs to ident
puteri [66]

Answer:b

Explanation: the particles will soon dissolve which means that the particles are changing but the volume is staying the same and the concentration is changing

3 0
2 years ago
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It's sunny in New York But it is cooL During the first part of the week, we might find this lizard basking in the sun
Leni [432]
Cold blooded animal? it’s a bit vague sorry
5 0
2 years ago
The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochlori
Dvinal [7]

Answer:

K(48.5°C) = 1.017 E-8 s-1

Explanation:

  • CH3Cl + H2O → CH3OH + HCl

at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1

at T2 = 48.5°C (321.5 K) ⇒ K2 = ?

Arrhenius eq:

  • K(T) = A e∧(-Ea/RT)
  • Ln K = Ln(A) - [(Ea/R)(1/T)]

∴ A: frecuency factor

∴ R = 8.314 E-3 KJ/K.mol

⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)

⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)

(1)/(2):

⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)

⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)

⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)

⇒ Ln (K1/K2) = - 3.422

⇒ K1/K2 = e∧(-3.422)

⇒ (3.32 E-10 s-1)/K2 = 0.0326

⇒ K2 = (3.32 E-10 s-1)/0.0326

⇒ K2 = 1.017 E-8 s-1

7 0
3 years ago
Which body system processes food into a useable source of energy?
anastassius [24]

Answer:

Glucose, found in the food animals eat, is broken down during the process of cellular respiration into an energy source called ATP. When excess ATP and glucose are present, the liver converts them into a molecule called glycogen, which is stored for later use.

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3 years ago
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