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Rasek [7]
3 years ago
5

Pls help me ASAP thank you!!

Chemistry
1 answer:
Stolb23 [73]3 years ago
7 0
Sorry i cant even read that
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Low melting point and poor conductor of electricity
WINSTONCH [101]

i THINK THE ANSWER IS A

5 0
3 years ago
What type of reaction is shown below?
natima [27]
An exothermic reaction is a type of chemical reaction in which energy is released to the environment in form of heat or light. Endothermic reaction in the other hand is a chemical reaction where energy is taken from the surroundings and thus the surroundings end up with less energy than they started with. In this case; the above reaction is an Exothermic reaction (heat is being released to the surroundings).
8 0
3 years ago
The activation energy of an uncatalyzed reaction is 95kJ/mol. The addition of a catalyst lowers the activation energy to 55kJ/mo
notka56 [123]

Answer:

a) at 25°C the rate of reaction increases by a factor of 1,027*10^7

b) at 25°C the rate of reaction increases by a factor of 1,777*10^5

Explanation:

using the Arrhenius equation

k= ko*e^(-Ea/RT)

where

k= reaction rate

ko= collision factor

Ea= activation energy

R= ideal gas constant= 8.314 J/mol*K

T= absolute temperature

for the uncatalysed reaction

k1= ko*e^(-Ea1/RT)

for the catalysed reaction

k2= ko*e^(-Ea2/RT)

dividing both equations

k2/k1= e^(-(Ea2-Ea1)/RT)

a) at 25°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,027*10^7

therefore at 25°C , k2/k1 = 1,027*10^6

b) at 125°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,777*10^5

therefore at 125°C , k2/k1 = 1,777*10^5

Note:

when the catalysts is incorporated, the catalysed reaction and the uncatalysed one run in parallel and therefore the real reaction rate is

k real = k1 + k2 = k2 (1+k1/k2)

since k2>>k1 → 1+k1/k2 ≈ 1 and thus k real ≈ k2

6 0
3 years ago
What masses of monobasic and dibasic sodium phosphate will you use to make 250 mL of 0.1 M sodium phosphate buffer, pH = 7?
Oduvanchick [21]

Answer:

  • Mass of monobasic sodium phosphate = 1.857 g
  • Mass of dibasic sodium phosphate = 1.352 g

Explanation:

<u>The equilibrium that takes place is:</u>

H₂PO₄⁻ ↔ HPO₄⁻² + H⁺    pka= 7.21 (we know this from literature)

To solve this problem we use the Henderson–Hasselbalch (<em>H-H</em>) equation:

pH = pka + log\frac{[A^{-} ]}{[HA]}

In this case [A⁻] is [HPO₄⁻²], [HA] is [H₂PO₄⁻], pH=7.0, and pka = 7.21

If we use put data in the <em>H-H </em>equation, and solve for [HPO₄⁻²], we're left with:

7.0=7.21+log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\ -0.21=log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\\\10^{-0.21} =\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\0.616 * [H2PO4^{-}] = [HPO4^{-2}]

From the problem, we know that [HPO₄⁻²] + [H₂PO₄⁻] = 0.1 M

We replace the value of [HPO₄⁻²] in this equation:

0.616 * [H₂PO₄⁻] + [H₂PO₄⁻] = 0.1 M

1.616 * [H₂PO₄⁻] = 0.1 M

[H₂PO₄⁻] = 0.0619 M

With the value of [H₂PO₄⁻]  we can calculate [HPO₄⁻²]:

[HPO₄⁻²] + 0.0619 M = 0.1 M

[HPO₄⁻²] = 0.0381 M

With the concentrations, the volume and the molecular weights, we can calculate the masses:

  • Molecular weight of monobasic sodium phosphate (NaH₂PO₄)= 120 g/mol.
  • Molecular weight of dibasic sodium phosphate (Na₂HPO₄)= 142 g/mol.

  • mass of NaH₂PO₄ = 0.0619 M * 0.250 L * 120 g/mol = 1.857 g
  • mass of Na₂HPO₄ = 0.0381 M * 0.250 L * 142 g/mol = 1.352 g
5 0
3 years ago
Where does energy come from? newsela
guapka [62]

Answer:

All of our energy comes from the sun.

Explanation:

The sun sends out huge amounts of energy through its rays every day. Without the sun, life on earth would not exist, since our planet would be totally frozen.

3 0
3 years ago
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