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PolarNik [594]
3 years ago
13

Protons have about the same mass as: A. electrons B. neutrons C. plasma particles

Chemistry
1 answer:
Andrews [41]3 years ago
3 0
Protons have the same mass as B. Neutrons. They have almost the same exact mass.
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A boy with pneumonia has lungs with a volume of 1.7 L that fill with 0.070 mol of air when he inhales. When he exhales, his lung
BlackZzzverrR [31]

Answer:

0.053moles

Explanation:

Hello,

To calculate the number of moles of gas remaining in his after he exhale, we'll have to use Avogadro's law which states that the volume of a given mass of gas is directly proportional to its number of moles provided that temperature and pressure are kept constant. Mathematically,

V = kN, k = V / N

V1 / N1 = V2 / N2= V3 / N3 = Vx / Nx

V1 = 1.7L

N1 = 0.070mol

V2 = 1.3L

N2 = ?

From the above equation,

V1 / N1 = V2 / N2

Make N2 the subject of formula

N2 = (N1 × V2) / V1

N2 = (0.07 × 1.3) / 1.7

N2 = 0.053mol

The number of moles of gas in his lungs when he exhale is 0.053 moles

7 0
3 years ago
A particular first-order reaction has a rate constant of 1.35 × 102 s-1 at 25.0°C. What is the magnitude of k at 95.0°C if Ea =
never [62]

Answer:

k ≈ 9,56x10³ s⁻¹

Explanation:

It is possible to solve this question using Arrhenius formula:

ln\frac{k2}{k1} = \frac{-Ea}{R} (\frac{1}{T2} -\frac{1}{T1} )

Where:

k1: 1,35x10² s⁻¹

T1: 25,0°C + 273,15 = 298,15K

Ea = 55,5 kJ/mol

R = 8,314472x10⁻³ kJ/molK

k2 : ???

T2: 95,0°C+ 273,15K = 368,15K

Solving:

ln\frac{k2}{k1} = 4,257

\frac{k2}{k1} = 70,593

{k2} = 9,53x10^3 s^{-1}

<em>k ≈ 9,56x10³ s⁻¹</em>

I hope it helps!

5 0
3 years ago
Which of the following is an example of a compound? A. helium B. water C. hydrogen B. oxygen
liq [111]

the compound is B.

Water is a compound because it is made up of water molecules.

7 0
2 years ago
Read 2 more answers
Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C
ivanzaharov [21]

The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O  (1)  

The enthalpy of reaction (1) is given by:

\Delta H = \Delta H_{r} - \Delta H_{p}   (2)

Where:

r: is for reactants

p: is for products

The bonds of the compounds of reaction (1) are:

  • 2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond
  • 7O₂: 7 moles of 1 O=O bond  
  • 4CO₂: 4 moles of 2 C=O bonds  
  • 6H₂O: 6 moles of 2 H-O bonds

Hence, the enthalpy of reaction (1) is (eq 2):

\Delta H = \Delta H_{r} - \Delta H_{p}

\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})      

\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})  

\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol  

\Delta H = -2860 kJ/mol          

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.

Read more here:

brainly.com/question/11753370?referrer=searchResults  

I hope it helps you!        

7 0
3 years ago
The percent abundance of each isotope tells you how many of each kind of isotope exist in every 100 particles. What does relativ
fiasKO [112]

Answer:

The answer is Relative plenitude alludes to the amount of a specific isotope is available in a given measure of test.  

Explanation:

The 'relative plenitude' of an isotope implies the level of that specific isotope that happens in nature. Most components are comprised of a blend of isotopes. The total of the rates of the particular isotopes must indicate 100%. The relative nuclear mass is the weighted normal of the isotopic masses. The percent plenitude of every sort of sweets reveals to you what number of every sort of Aufbau there are in each 100 CANDIES. Percent wealth is additionally relative plenitude. This is only a method for giving us a photo on which kind exists all the more every now and again.

3 0
3 years ago
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