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Lesechka [4]
3 years ago
7

If you have 65.8 grams NH3, how many grams of F2 are required for complete reaction?

Chemistry
1 answer:
Mashcka [7]3 years ago
6 0

Answer:75

Explanation:Hello Bella, how are you today? Looking for my answer I see. Good luck, and next time don't use your real name.

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Combustion analysis of 0.300 g of an unknown compound containing carbon, hydrogen, and oxygen produced 0.5213 g of co2 and 0.283
Lyrx [107]
First, we have to get how many grams of C & H & O in the compound:
- the mass of C on CO2 = mass of CO2*molar mass of C /molar mass of CO2
                                        = 0.5213 * 12 / 44 = 0.142 g
- the mass of H atom on H2O = mass of H2O*molar mass of H / molar mass of H2O
                                                 =0.2835 * 2 / 18 = 0.0315 g
- the mass of O = the total mass - the mass of C atom - the mass of H atom
                          =  0.3 - 0.142 - 0.0315 = 0.1265 g
Convert the mass to mole by divided by molar mass
C(0.142/12) H(0.0315/2) O(0.1265/16) 
C(0.0118) H(0.01575) O(0.0079) by dividing by the smallest value 0.0079
C1.504 H3.99 O1 by rounding to the nearst fraction
C3/2 H4/1 )1/1 multiply by 2 
∴ the emprical formula C3H8O2
           



6 0
3 years ago
5.5 cm east + 3.5 cm west =
Firlakuza [10]
2 cm east is the answer because you go east 5.5 cm and then go back in a sense 3.5 so its basically 5.5-3.5 because its backwards while still facing eat
4 0
3 years ago
Which best describes the tyndall effect? the scattering of light by solutes in a mixture the scattering of light by solvent in a
Oksanka [162]

the scattering of light by particles in a mixture

3 0
3 years ago
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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
morpeh [17]

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

6 0
3 years ago
A catalyst
kotegsom [21]
A catalyst speeds up the rate of reaction so the answer is B.
8 0
3 years ago
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