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4 years ago

A metal surface becomes dull because of continued abrasion physical change or chemical

Chemistry

1 answer:

Evgesh-ka [11]4 years ago

6 0

It had to be a physical change unless there a chemical touching the metal surface making it a chemical change .

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The formation of tert-butanol is described by the following chemical equation:

fomenos

Answer:

Second step

(CH3)3C+ (aq) + OH^-(aq) ------->(CH3)3COH(aq)

Explanation:

This reaction involves;

First the ionization of the tertiary halide to firm a carbocation

Secondly the attack of the hydroxide ion on the carbocation to form tert-butanol

First step;

(CH3)3CBr (aq) → (CH3)3C+ (aq) + Br- (aq)

Second step

(CH3)3C+ (aq) + OH^-(aq) ------->(CH3)3COH(aq)

This second step completes the reaction mechanism.

3 0

3 years ago

Need help fast please 20 points

artcher [175]

Answer:

gamma radiation

Explanation:

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5 0

3 years ago

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You have four bottles labeled 1, 2, 3, 4. Each bottle has a white powder in it. The identities of the substances are held secret

KATRIN_1 [288]

If you are using usatestprep its bottle 2

4 0

4 years ago

Read 2 more answers

What is the hydronium ion concentration of a solution whose pH is 7.30

Assoli18 [71]

[ H₃O⁺] = 10 ^ - pH

[ H₃O⁺ ] = 10 ^ - 7.30

[ H₃O⁺ ] = 5.011 x 10⁻⁸ M

hope this helps!

6 0

3 years ago

Which statement below correctly describes the relationship between Q and K for both reactions? Are these reactions spontaneous a

nikklg [1K]

Answer:

Q < K for both reactions. Both are spontaneous at those concentrations of substrate and product.

Explanation:

Hello,

In this case, the undergoing chemical reactions with their proper Gibbs free energy of reaction are:

$A->B;\Delta _rG^o=-13 kJ/mol$

$C ->D ;\Delta _rG^o=3.5 kJ/mol$

The cellular concentrations are as follows: [A] = 0.050 mM, [B] = 4.0 mM, [C] = 0.060 mM and [D] = 0.010 mM.

For each case, the reaction quotient is:

$Q_1=\frac{4.0mM}{0.050mM}=80\\ Q_2=\frac{0.010mM}{0.060mM}=0.167$

A typical temperature at a cell is about 30°C, in such a way, the equilibrium constants are:

$K_1=exp(-\frac{-13000J/mol}{8.314J/mol*K*303.15K} )=173.8\\K_2=exp(-\frac{3500J/mol}{8.314J/mol*K*303.15K} )=0.249$

Therefore, Q < K for both reactions. Both are spontaneous at those concentrations of substrate and product.

Best regards.

6 0

3 years ago