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Paha777 [63]
2 years ago
7

Can someone convert 150 cm of copper (Cu) wire into mm of Cu wire?​

Chemistry
1 answer:
Vedmedyk [2.9K]2 years ago
6 0

1500 mm of copper wire are produced by converting 150 cm of copper wire.

The following is the centimeter to millimeter conversion factor: 1 cm is equivalent to 10 mm.

Consequently, 150 cm will equal 150 x 10 = 1500 mm.

  • The multi-step process of unit conversion involves multiplying or dividing by a numerical factor.
  • Unit conversion is the process of converting the measurement of a given amount between various units, often by multiplicative conversion factors that alter the value of the measured quantity without altering its effects.
  • Unit conversion is a multi-step procedure that involves adding, subtracting, multiplying, or dividing by a conversion factor.
  • The process may also require the selection of the correct number of significant digits, and rounding.

To learn more about unit conversion visit:

brainly.com/question/11543684

#SPJ9

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Who was given name black hole​
marin [14]

Answer:

John Archibald Wheeler

3 0
3 years ago
Write down the formula for B<br> example:<br> Hydrogen + Fluorine = Hydrogen Fluorine <br><br> help
Tom [10]

Answer:

P³⁻ + Cl⁻ --> PCl₃

Explanation:

PCl₃: phosphorus trichloride. prefix in front of chloride is "tri"–meaning three.

5 0
3 years ago
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How many grams of mercury can be produced if 18.0 g of mercury (11) oxide decomposes?
NARA [144]

Answer:

18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury

Explanation:

Mercury oxide has molar mass of 216.6 g/ mol. It gas a molecular formula of HgO.

The decomposition of mercury oxide is given by the chemical equation below:

2HgO ----> 2Hg + O₂

2 moles of HgO decomposes to produce 1 mole of Hg

2 moles of HgO has a mass of 433.2 g

433.2 g of HgO produces 216.6 g of Hg

18.0 of HgO will produce 18 × 216.6/433.2 g of Hg = 9.0 g of Hg

Therefore, 18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury

3 0
3 years ago
What is the factor of 1024
Ugo [173]
1
2
4
8
16
32
64
128
256
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1024
4 0
3 years ago
Suppose you were preparing 1.0 L of a bleaching solution in a volumetric flask, and it calls for 0.21 mol of NaOCl. If all you h
yulyashka [42]

Answer:

0.256 L  

Explanation:

We should use the following formula:

concentration (1) × volume (1) =  concentration (2) × volume (2)

concentration (1) = 0.82 M NaOCl

volume (1) = ?

concentration (2) = 0.21 M NaOCl

volume (2) = 1 L

volume (1) = [concentration (2) × volume (2)] / concentration (1)

volume (1) = [0.21 / 1] / 0.82 = 0.256 L

3 0
3 years ago
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