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Katena32 [7]
3 years ago
9

Bob, Jill, Kim, and Steve measure an object's length, density, mass, and brightness, respectively. Which student must derive a u

nit of measurement?
Jill
Bob
Kim
Steve
Chemistry
1 answer:
ValentinkaMS [17]3 years ago
8 0

Answer:

The answer is Bob

Explanation:

The science exist to kinds of measurements base units and derive units of measurement.

Base units are the ones that measure: length, mass, time, electric current, temperature, luminous intensity and amount of substance.

Derivative units of measurement are obtain when we combine the base units.

Know from your question

Bob measures length is a base unit

Jill measures density is a derivative unit, is the combination of mass and volumen

Kim measures mass is a base unit

Steve measures brightness is a base unit.

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What is the penetrating power of beta
zlopas [31]
The penetrating power of alpha rays, beta rays, and gamma rays varies greatly. Alpha particles can be blocked by a few pieces of paper. Beta particles pass through paper but are stopped by aluminum foil. Gamma rays are the most difficult to stop and require concrete, lead, or other heavy shielding to block them.
3 0
3 years ago
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
Need help with this can some please help
vekshin1
I believe you would just put a 2 in front of NH3 and keep the other ones as 1
7 0
3 years ago
7.66 Write balanced equations for the following reactions: (a) potassium oxide with water, (b) diphosphorus trioxide with water,
White raven [17]

Explanation:

(a) potassium oxide with water

K_2O(s)+H_2O(l)\rightarrow 2KOH(aq)

According to reaction,1 mole of potassium  oxide reacts with 1 mole of water to give 1 mole of potassium hydroxide.

(b) diphosphorus trioxide with water

P_2O_3(s)+3H_2O(l)\rightarrow 2H_3PO_3(aq)

According to reaction,1 mole of diphosphorus trioxide reacts with 2 moles of water  to give 2 moles of phosphorus acid.

(c) chromium(III) oxide with dilute hydrochloric acid,

Cr_2O_3(s)+6HCl(aq)\rightarrow 2CrCl_3(aq)+3H_2O(l)

According to reaction,1 mole of chromium(III) oxide reacts with 6 moles of hydrochloric acid to give 2 moles of chromium(III) chloride and 3 moles of water.

(d) selenium dioxide with aqueous potassium hydroxide

SeO_2 (s)+2KOH (aq)\rightarrow K_2SeO_3(aq)+H_2O(l)

According to reaction,1 mole of selenium dioxide reacts with 2 moles of potassium hydroxide to give 1 mole of potassium selenite and 1 mole of water.

6 0
3 years ago
Write the name of the main gases in the air​
Delicious77 [7]

Answer:

Nitrogen

Oxygen

Argon

Carbon Dioxide

Methane

Ozone

Explanation:

N₂ accounts for 78% of the atmosphere.

O₂ accounts for 21% of the atmosphere.

Ar accounts for 0.9% of the atmosphere.

CO₂, CH₄, and O₃ only take up 0.1% of the atmosphere.

3 0
3 years ago
Read 2 more answers
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