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Zepler [3.9K]
3 years ago
13

Please answer correctly. No links please. How many pounds are there in 1.53 tons

Chemistry
1 answer:
Lana71 [14]3 years ago
6 0

Answer:

3060 lbs

Explanation:

multiply mass by 2000

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A graduate cylinder was filled to 25.0 mL with a liquid. A solid object weighting 39.7 g was immersed in the liquid, raising the
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3 years ago
If a temperature increase from 21.0 ∘c to 35.0 ∘c triples the rate constant for a reaction, what is the value of the activation
Airida [17]

Answer:

59.077 kJ/mol.

Explanation:

  • From Arrhenius law: <em>K = Ae(-Ea/RT)</em>

where, K is the rate constant of the reaction.

A is the Arrhenius factor.

Ea is the activation energy.

R is the general gas constant.

T is the temperature.

  • At different temperatures:

<em>ln(k₂/k₁) = Ea/R [(T₂-T₁)/(T₁T₂)]</em>

k₂ = 3k₁ , Ea = ??? J/mol, R = 8.314 J/mol.K, T₁ = 294.0 K, T₂ = 308.0 K.

ln(3k₁/k₁) = (Ea / 8.314 J/mol.K) [(308.0 K - 294.0 K) / (294.0 K x 308.0 K)]

∴ ln(3) = 1.859 x 10⁻⁵ Ea

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4 0
3 years ago
When 5.00 g of NH4NO3 is dissolved in 100.0 g of water in a styrofoam cup, the T = - 3.82 ^°C. Is the process of dissolving ammo
miv72 [106K]

Answer:

The process is endothermic.

Heat is 26,9 kJ/mol

Explanation:

The dissolution of NH₄NO₃ in water is:

NH₄NO₃ → NH₄⁺ + NO₃⁻

As the change of temperature in the cup is ΔT = -3,82°C

<em>The process is endothermic</em>. Because the temperature is decreasing in the process. That means the process needs heat.

Assuming the  heat capacity of the solution is 4.18 J/Kg :

q = -C×m×ΔT

Where q is heat, C is heat capacity (4,18J/Kg), m is mass (105g) and ΔT is change in temperature (-3,82°C)

q = 1677 J ≈ <em><u>1,68kJ</u></em>

The moles of NH₄NO₃ dissolved are:

5,00g × (1mol / 80,043g) = <em><u>0,0625 moles</u></em>

That means heat of this process in kJ/ mol is:

1,68kJ / 0,0625moles = <em>26,9 kJ/mol</em>

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I hope it helps!

8 0
3 years ago
Predict the products of the reaction below. that is, complete the right-hand side of the chemical equation. be sure your equatio
kkurt [141]
The equation is as follow,

<span>                                  HBr </span>₍aq₎  +  H₂O ₍l₎    →

Solution:
             HBr being strong acid with Ka value of 1.0 × 10⁹. When HBr is added to water, water acts as a base and HBr acts as a acid. Water picks the proton (H⁺) from HBr and converts into Conjugate acid (H₃O⁺) ahile HBr is converted into Conjugate Base (Br⁻) after loosing proton. The equation for this reaction is as follow,

                      HBr ₍aq₎  +  H₂O ₍l₎    →    H₃O⁺ ₍aq₎  +  Br⁻ ₍aq₎
8 0
4 years ago
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