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Viefleur [7K]
3 years ago
7

What is 24,506 written in scientific notation?

Chemistry
1 answer:
irinina [24]3 years ago
6 0

Answer:

2.4506 x 10^{4}

Explanation:

move decimal point to the right 1 time : 24506

24506 becomes 2.4506

count how many numbers after decimal point and that becomes b

b = 4

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Because it is less stable than phosphorous
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Describing the Rock Cycle
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Answer:

A, C, and D

Explanation:

A-Any type of rock can change into any other type of rock by weathering and erosion.

C-Rocks change slowly over time.

D- The rock cycle shows how the three rock types relate to one another.

Its correct on edge

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KBr + AgNO3 → AgBr + KNO3<br><br><br> Which two chemicals are products in the chemical reaction?
iragen [17]

Answer:

Products are AgBr and KNO3

4 0
3 years ago
A sealed helium balloon with an internal pressure of 1.00 atm and a volume of 4.50 L at 293 K is moved. What volume will the bal
bixtya [17]

Answer:

6.48 L

Explanation:

From the question,

Applying

PV/T = P'V'/T'......................... Equation 1

P = initial pressure of the helium balloon, V =  Initial volume of the balloon, T = Initial temperature of the balloon, P' = Final pressure of the balloon, T' = Final temperature of the balloon, V' = Final volume of the balloon.

make V' the subject of the equation

V' = PVT'/P'T......................... Equation 2

Given: P = 1 atm, V = 4.5 L, T' = 253 K, T= 293 K, P' = 0.6 atm

Substitute these values into equation 2

V' = (4.5×1×253)/(0.6×293)

V' = 1138.5/175.8

V' = 6.48 L

8 0
2 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
2 years ago
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