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SVEN [57.7K]
3 years ago
11

The hypotenuse of a right triangle is 5 m long. The longer leg is 1 m longer than the shorter leg. Find the side lengths of the

triangle.
Mathematics
1 answer:
algol [13]3 years ago
7 0
The shortest side is 3 m and the other side is 4 m. If you put all these numbers in the Pythagorean Theorem it checks out. (3^2+4^2=5^2)
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54 divided by four is 13.5 or 27/2
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Solve –15 = 4m – 7.<br><br><br> –32<br><br> 2<br><br> –12<br><br> –2
Shkiper50 [21]

Answer:

-2 =m

Step-by-step explanation:

–15 = 4m – 7

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3 years ago
Is 9/20 closer to 0 , 1/2 or 1
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Answer:

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When electricity (the flow of electrons) is passed through a solution, it causes an oxidation-reduction (redox) reaction to occu
oee [108]

Answer:

a. 135 g

b. 60.6 min

Step-by-step explanation:

a. What mass of Cu(s) is electroplated by running 28.5 A of current through a Cu2+ (aq) solution for 4.00 h? Express your answer to three significant figures and include the appropriate units.

The chemical equation for the reaction is given below

Cu²⁺(aq) + 2e⁻ → Cu(s)

We find the number of moles of Cu that are deposited from

nF = It where n = number of moles of electrons, F = Faraday's constant = 96485 C/mol, I = current = 28.5 A and t = time = 4.00 h = 4.00 × 60 min/h × 60 s/min = ‭14,400‬ s

So, n = It/F = 28.5 A × ‭14,400‬ s/96485 C/mol = ‭410,400‬ C/96485 C/mol = 4.254 mol

Since 2 moles of electrons deposits 1 mol of Cu, then 4.254 mol of electrons deposits 4.254 mol × 1 mol of Cu/2 mol = 2.127 mol of Cu

Now, number of moles of Cu = n' = m/M where m = mass of copper and M = molar mass of Cu = 63.546 g/mol

So, m = n'M

= 2.127 mol × 63.546 g/mol

= 135.15 g

≅ 135 g to 3 significant figures

b. How many minutes will it take to electroplate 37.1 g of gold by running 5.00 A of current through a solution of Au+(aq)?

The chemical equation for the reaction is given below

Au⁺(aq) + e⁻ → Au(s)

We need to find the number of moles of Au in 37.1 g

So, number of moles of Au = n = m/M where m = mass of gold = 37.1 g and M = molar mass of Au = 196.97 g/mol

So, n = m/M = 37.1 g/196.97 g/mol = 0.188 mol

Since 1 mol of Au is deposited  by 1 moles of electrons, then 0.188 mol of Au deposits 0.188 mol of Au × 1 mol of electrons/1 mol of Au = 0.188 mol of electrons

We find the time it takes to deposit 0.188 mol of electrons that are deposited from

nF = It where n = number of moles of electrons, F = Faraday's constant = 96485 C/mol, I = current = 5.00 A and t = time

So, t = nF/It

= 0.188 mol × 96485 C/mol ÷ 5.00 A

= ‭18173.30‬ C/5.00 A

= 3634.66 s

= 3634.66 s × 1min/60 s

= 60.58 min

≅ 60.6 min to 3 significant figures

6 0
3 years ago
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