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hodyreva [135]
3 years ago
7

An atomega has a mass of 53 an an atomic number of 26. How many neutrons does it have

Chemistry
1 answer:
Ghella [55]3 years ago
4 0
26, protons and nuetrons will always be the  same

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Both hydrogen sulfide (H2S) and ammonia (NH3)
faust18 [17]

Answer: The molar mass of H2S is greater than the molar mass of NH3, making the velocity and effusion rate of NH3 particles faster.

Effusion rate is inversely proportional to molar mass.

NH3 will have a higher average molecule velocity, so it will diffuse faster and will reach the other side of the room more quickly.

Explanation: change up your response a bit

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3 years ago
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What would be an advantage of using a solar cell as a model for a leaf?
Sidana [21]
A solar cell has a similar function to a leaf.

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3 years ago
If a solution of HF (Ka=6.8×10−4) has a pH of 2.90, calculate the concentration of hydrofluoric acid.
fgiga [73]

Answer: 0.0023

Explanation:

I just answered this very same question. You can read my answer here: brainly.com/question/12077289.

1) First, you must write the balanced equilibrium equation, which will let you to determine the mole ratios of the different species in solution:

    HF ⇆ H⁺ + F⁻

Therefore, the mole ratio is: 1 HF: 1 H⁺ : 1  F⁻

2) Second, write the equilibrium constant of the acid, Ka, which will permit you to state the relationship of the concentrations in equilibrium:

        Ka = [H⁺] [F⁻] / [HF]

   [HF] is the searched concentration of the hydrofluoric acid

   From the mole ratio of the balanced chemical equation [H⁺] = [F⁻]

       Hence:

       Ka = [H⁺] [H⁺] / [HF] = [H⁺]² / [HF]

3) From the pH value you can calculate [H⁺],  using the definition:

  •    pH= - log [H⁺]
  •    Substitute: 2.90 = - log [H⁺]

 

 Clear [H⁺] using logarithm properties:

         [H^+]=10^{-2.90}=0.00126

4) Now you can substitute [H⁺] and the value of Ka in the equation for Ka:

   Ka = [H⁺]² / [HF] ⇒ [HF] = [H⁺]² / Ka = (0.00126)² / (6.8×10⁻⁴) = 0.00233.

Since the value of Ka has two significant figures, you must report the answer with two significan figures, i.e. 0.0023

7 0
3 years ago
Take 3 drops per 15 lbs. of body weight per day divided into 6 doses until the cooties are gone. how many drops do you take per
Luden [163]
You'd take about ten or eleven drops per day
3 0
3 years ago
Calculate the heat required to melt 7.35 g of benzene at its normal melting point. Heat of fusion (benzene) = 9.92 kJ/mol Heat =
grigory [225]

Answer:

The heat required to melt 7.35 g of benzene at its normal melting point is 934.8 Joules.

The heat required to vaporize 7.35 g of benzene at its normal melting point is 2,893 Joules.

Explanation:

Mass of benzene = 7.35 g

Moles of benzene = \frac{7.35 g}{78 g/mol}=0.09423 mol

Heat fusion of benzene,\Delta H_{fus} = 9.92 kJ/mol

1) Heat required to melt 7.35 g of benzene at its normal melting point = Q

Q=\Delta H_{fus}\times 0.09423 mol

=9.92 kJ/mol\times 0.09423 mol=0.9348 kJ=934.8 J

(1 kJ = 1000 J)

2) Heat vaporization of benzene,\Delta H_{vap} = 30.7 kJ/mol

Heat required to vaporize 7.35 g of benzene at its normal melting point = Q

Q=\Delta H_{Vap}\times 0.09423 mol

=30.7 kJ/mol\times 0.09423 mol=2.893 kJ=2,993 J

(1 kJ = 1000 J)

3 0
3 years ago
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