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ANTONII [103]
3 years ago
7

What's 1.8499 x 10^9 in standard notation

Mathematics
1 answer:
maxonik [38]3 years ago
5 0
 you just have to move the decimal point 9 places to the right thats an easy trick to remember 1849900000


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Y-3=-3y-43 how would i solve?​
frez [133]

Answer: y−3=−3y−43

Step 1: Add 3y to both sides.

y−3+3y=−3y−43+3y

4y−3=−43

Step 2: Add 3 to both sides.

4y−3+3=−43+3

4y=−40

Step 3: Divide both sides by 4.

4y

4

=

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y=−10

Answer:

y=−10

Step-by-step explanation: brainliest:)

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3 years ago
Sophie went to the lake with her family. First, they spent 1 1/4 hours of fishing. Then, they spent twice as much time swimming.
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Overall, they spent 3 3/4 hours fishing and swimming.
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3 years ago
A car takes 54.9 seconds to travel 1 mile. How long does it take the car to travel 4.3 miles? First round to the nearest whole n
fiasKO [112]

Answer: Rounded: 236 seconds

Not rounded: 236.07 seconds

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3 years ago
24 . What is the length of Side Of BC <br> A 3<br> C 10<br> B 8<br> D 24
jekas [21]

Answer:

B (8)

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AB = BC = AC

AB = BC

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3 years ago
Read 2 more answers
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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