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melisa1 [442]
4 years ago
6

Convert the speed of light, 3.0 × 10^8 m/sec, into mi/hr

Chemistry
1 answer:
Phantasy [73]4 years ago
4 0
[Hint: 1 m/s=2.237 mi/hr] Write it out to get an understanding of how to convert. 1} 3.0x10^8m/s over 1. (Don't forget to put your units as stated above) 2} Utilizing the hint given you want to cancel out the unit that is being converted. (Ex: m/s) -> 3.0x10^8 m/s over 1 TIMES 2.237 mi/hr over 1 m/s Now notice that the m/s is on top and on bottom now, if you've written it out. The numerator has something in common with the denominator and can now be canceled out. (Cross out the m/s NOT the digits JUST the units) 3} You have now narrowed it down to digits and the desired unit you want. -> 3.0x10^8 over 1 TIMES 2.237 mi/hr over 1 Remember: multiply the numbers straight across the numerator to get a new total for the numerator. The same will go for the denominator. Once you have 1 fraction left, it basically becomes a math equation. Num ■(3.0×10^8)(2.237)= 671100000 Den ■ (1)(1)= 1 4} 671100000(mi/hr)/1= 671100000 mi/hr If you thought you were done you're wrong. You have to consider the significant figures or sig figs. Learn the rules to sig figs and you'll ace this part. You're answer is going to be 3 sig figs. Therefore, your answer is: 6.71x10^8 mi/hr
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Calculate the standard potential for the following galvanic cell: Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s) which has the overall bala
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Answer:

1.06  V  

Explanation:

The standard reduction potentials are:

Ag^+/Ag     E° =  0.7996 V  

Ni^2+/Ni     E° = -0.257   V

The half-cell and cell reactions for Ni | Ni^2+ || Ag^+ | Ag are

Ni → Ni^2+ + 2e-                     E° = 0.257   V

<u>2Ag^+ 2e- → 2Ag               </u>    <u>E° = 0.7996 V </u>

Ni + 2Ag^+ → Ni^2+ + 2Ag     E° = 1.0566  V

To three significant figures, the standard potential for the cell is 1.06 V .

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4 years ago
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Most elements are _____? <br><br> nonmetals <br> metalloids<br> metals
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4 0
2 years ago
6 CO2 + 12 H2O C6H12O6 + 6 H2O + 6 O2<br><br>how many hydrogen atoms are involved in this reaction?
Romashka-Z-Leto [24]
How many hydrogen atoms are involved in this reaction? 3
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4 years ago
A mixture of calcium carbonate, CaCO3, and barium carbonate, BaCO3, weighing 5.40 g reacts fully with hydrochloric acid, HCl(aq)
amid [387]

Answer:

CaCO₃ = 85.18%

BaCO₃ = 14.82%

Explanation:

The acid will react with the salts, the partial reactions are:

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

BaCO₃ + 2HCl → BaCl₂ + CO₂ + H₂O

So, the total amount of CaCO₃ BaCO₃ will form the CO₂.

Using the ideal gas law to calculate the number of moles of CO₂:

PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.082 atm*L/mol*K), and T is the temperature (50ºC + 273 = 323 K).

0.904*1.39 = n*0.082*323

26.486n = 1.25656

n = 0.05 mol

So, the number of moles of the mixture is 0.05 mol.

The molar masses of the components are:

CaCO₃ = 40 g/mol of Ca + 12 g/mol of C + 3*16 g/mol of O = 100 g/mol

BaCO₃ = 137.3 g/mol of Ba + 12 g/mol of C + 3*16 g/mol of O = 197.3 g/mol

Let's call x the number of moles of CaCO₃ and y the number of moles of BaCO₃, so:

100x + 197.3y = 5.4

x + y = 0.05 mol

y = 0.05 - x

100x + 197.3*(0.05 - x) = 5.4

100x - 197.3x = 5.4 - 9.865

97.3x = 4.465

x = 0.046 mol of CaCO₃

y = 0.004 mol of BaCO₃

So, the masses are:

CaCO₃ = 100* 0.046 = 4.60 g

BaCO₃ = 137.3*0.004 = 0.80 g

The percentages in the mixture are:

CaCO₃ = (4.60/5.40)*100% = 85.18%

BaCO₃ = (0.80/5.40)*100% = 14.82%

4 0
3 years ago
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