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3 years ago

I do not have a question anymore

Chemistry

2 answers:

vlabodo [156]3 years ago

6 0

Is this supposed to be for free poi nts??

IrinaK [193]3 years ago

4 0

Answer:

I dont have a answer

Explanation:

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How will designing more efficient stoves help conserve fuel?

Anastasy [175]

Answer:

it would do wonders for the health of those using them and, by extracting energy from the fuel more efficiently, would be more environmentally sustainable too.

5 0

2 years ago

Calculate the KE of an object that has a mass of 25kg that is moving at 5m/s2.

svp [43]

Answer:

312.5J

Explanation:

K.E=1/2*mv°2

=1/2*25*25

=312.5

4 0

4 years ago

Which of the slightly soluble salts below will be more soluble in acidic solution than in pure water?

AlexFokin [52]

Answer:

h. both Mg(OH)₂ and CaCO₃

Explanation:

Let's consider the solution of Mg(OH)₂ according to the following equation:

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

In acidic solution, OH⁻ reacts with H⁺ to form H₂O.

OH⁻(aq) + H⁺(aq) ⇄ H₂O(l)

According to Le Chatelier's principle, since [OH⁻] decreases, the solution of Mg(OH)₂(s) shifts toward the right, increasing its solubility.

Let's consider the solution of CaCO₃ according to the following equation:

CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)

In acidic solution, CO₃²⁻ reacts with H⁺ to form HCO₃⁻.

CO₃²⁻(aq) + H⁺(aq) ⇄ HCO₃⁻(aq)

According to Le Chatelier's principle, since [CO₃²⁻] decreases, the solution of CaCO₃(s) shifts toward the right, increasing its solubility.

Let's consider the solution of AgCl according to the following equation:

AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

Cl⁻ does not react with H⁺ because it comes from a strong acid (HCl). Therefore, the solubility of AgCl(s) is not affected by the pH.

6 0

3 years ago

Please help thank you!

insens350 [35]

South east should be the answer

7 0

3 years ago

The first-order isomerization reaction: cyclopropane → propene has a rate constant of 1.10×10-4 s-1 at 470°C and 5.70×10-4 s-1 a

BigorU [14]

Answer:

Activation energy, Ea, for the reaction 261.7 kj/mole

Explanation:

Given

Rate constant at 470 ⁰C (K₁) = 1.10 x 10⁻⁴ s⁻¹ and rate constant at 500⁰C (K₂) = 5.7 x 10⁻⁴ s⁻¹

Temperature (T₁) = 470 + 273 = 743 K and

temperature (T₂) = 500 + 273 = 773 K

Activation energy (Eₐ) = ?

Universal gas constant (R) = 8.314 J. K⁻¹. mole⁻¹

We know log $\frac{K_{2} }{K_{1} }$ = $\frac{E_{a} }{2.303XR}$ $\frac{(T_{2}-T_{1} ) }{T_{2}T_{1} }$

⇒ log $\frac{5.7 X 10^{-4} }{1.1 X 10^{-4} }$ = $\frac{E_{a} }{2.303X8.314}$ $\frac{(773 - 743)}{773X743}$

⇒ 0.714 = $\frac{E_{a}X30 }{10996950.3}$

⇒ Eₐ = $\frac{7851822.5}{30}$ j/mole = 261.7 Kj/mole

5 0

3 years ago