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Fiesta28 [93]
3 years ago
7

Must show units and how they cancelli 1.) 175 km to um 3.) 385 nm to dm 5.) 492 um tom 7.) 52 x 103 dm to mm 9.) 321x 1035 mm to

km 11.) 456 x 103 m to km 13.) 422 x 103 m to nm 15.) 4.87 x 1030m to pm 17.) 5.26 x 103 m to um 19.) 1.25 x 1035 m to Mm 21.) 4.22 x 103 Tm to nm
Chemistry
1 answer:
morpeh [17]3 years ago
6 0

Explanation:

1.) 175 km to μm

1 km=10^9 \mu m

175 km=175\times 10^9\mu m=1.75\times 10^{11} \mu m

3.) 385 nm to dm

1 nm=10^{-8} dm

385 nm=385\times 10^{-8} dm=3.85\times 10^{-6} dm

5.) 492 μm  to m

1 μm =  10^{-6} m

492 \μm=492\times 10^{-6} m=4.92\times 10^{-4} m

7.) 52\times 10^3 dm to mm

1 dm = 100 mm

52\times 10^3 dm=52\times 10^3\times 100 mm=5.2\times 10^{6}dm

9.) 321\times 10^{35} mm to km

1 mm = 10^{-6} km

321\times 10^{35} mm=321\times 10^{35}\times 10^{-6} km=3.21\times 10^{31} km

11.) 456\times 10^3 m to km

m = 0.001 km

456\times 10^3m =456\times 10^3 m\times 0.001 km=456 km

13.) 422\times 10^3 m to nm

1 m = 10^{9} nm

422\times 10^3 m=422\times 10^3\times 10^{9} nm=4.22\times 10^{14} nm

15.) 4.87\times 10^{30} m to pm

1 m = 10^{12} pm

4.87\times 10^{30} m=4.87\times 10^{30}\times 10^{12} pm=4.82\times 10^{42} pm

17.) 5.26\times 10^3 m to um

1 m =  10^{6} \mu m

5.26\times 10^3 m=5.26\times 10^3\times 10^6 \mu m=5.26\times 10^{9} \mu m

19.) 1.25\times 10^{35}m to Mm

1 m =  10^{-6} Mm

1.25\times 10^{35} m=1.25\times 10^{35}\times 10^{-6} Mm=1.25\times 10^{-29} Mm

21.) 4.22\times 10^3 Tm to nm

1 Tm = 10^{21} nm

4.22\times 10^3 Tm=4.22\times 10^3\times 10^{21} nm=4.22\times 10^{24} nm

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Answer:

136

Explanation:

The Mass Number is the combination of the amount of Protons and Neutrons in an element, so if the total mass is 222, and the amount of protons is 86, then you can do 86 + x = 222 to find that x is equal to 136

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In the synthesis reaction, the white powder produced is magnesium oxide (mgo, mm = 40.3 g/mol). how many moles of magnesium oxid
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Number of moles = mass of product / molecular mass
=mass of product (MgO) / 40.3

Since the mass of MgO is not given in the question, the correct answer choice cannot be given.  However, proceeding witht eh above formula will enable you to find the correct number of moles given the mass of MgO.
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A 500 gram piece of metal had a volume of 2.75cm3. What is it’s density
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What is the theoretical yield of fluorenone if you oxidize 175 mg of fluorene?
My name is Ann [436]

Answer:

  • 602 mg of CO₂ and 94.8 mg of H₂O

Explanation:

The<em> yield</em> is measured by the amount of each product produced by the reaction.

The chemical formula of <em>fluorene</em> is C₁₃H₁₀, and its molar mass is 166.223 g/mol.

The <em>oxidation</em>, also know as combustion, of this hydrocarbon is represented by the following balanced chemical equation:

        2C_{13}H_{10}+31O_2\rightarrow 26CO_2+10H_2O

To calculate the yield follow these steps:

<u>1. Mole ratio</u>

          2molC_{13}H_{10}:31molO_2:26molCO_2:10molH_2O

<u />

<u>2. Convert 175mg of fluorene to number of moles</u>

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  • Number of moles = mass in grams / molar mass

  • \text{number of moles}=0.175g/166.223g/mol=0.0010528mol

<u>3. Set a proportion for each product of the reaction</u>

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i) number of moles

         2molC_{10}H_{13}/26molCO_2=0.0010528molC_{10}H{13}/x

x=0.0010528molC_{10}H_{13}\times 26molCO_2/2molC_{10}H_{13}=0.013686molCO_2

ii) mass in grams

The molar mass of CO₂ is 44.01g/mol

  • mass = number of moles × molar mass
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i) number of moles

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ii) mass in grams

The molar mass of H₂O is 18.015g/mol

  • mass = number of moles × molar mass
  • mass = 0.00526 moles × 18.015 g/mol = 0.0948mg = 94.8 mg
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The maximum number of covalent bonds formed by a single carbon atom is 4

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