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avanturin [10]
3 years ago
15

A 10 gram sample of material, a, was placed in water. four grams of it, b, did not dissolve, even after being separated and plac

ed in a second volume of water. bcould not be decomposed into any simpler substances. evaporation of the water from the soluble portion of ayielded 6 grams of solid,
c. heating bin air yielded 8 grams of gas, d, leaving no solid residue. melting and electrolysis of cyielded 2.4 grams of metal, e, and 3.6 grams of gas, f. classify each material described above by placing the corresponding letter in the appropriate place in the table below.
Chemistry
1 answer:
kicyunya [14]3 years ago
3 0
Sksksk isjsjs jsjsjs jsjsjs jsjsjs jsjsjs ksjss
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Calculate the specific heat of the substance if 373 J is required to raise the temperature of a 312 gram sample by 25°C?
I am Lyosha [343]

Explanation:

Q = 373J

∆∅ = 25°C

m = 312g

c = x

Q = mc∆∅

373 = 312(x)(25)

373 = 7800x

x = 373/7800

x = 0.0478J/(g°C)

4 0
3 years ago
15. The starting diol for this molecule is
victus00 [196]

<h2>Answer with explanation </h2>

<h3><em>The starting diol for this molecule is :-</em></h3><h3><em>The starting diol for this molecule is :-D) ethan-1,2-diol.</em></h3>

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3 0
3 years ago
Arrange the following molecules in order of increasing bond polarity (highest bond polarity at the bottom):
Tom [10]
NF3– 0.94– third
NCl3–0.12– second
NBr3–0.08– first
CF4–1.43– fourth

NBr3—NCl3—NF3—CF4
Lowest. Highest
4 0
3 years ago
Magnesium occurs in seawater to the extent of 1.4 g magnesium per kilogram of seawater. What volume of seawater, in cubic meters
NISA [10]

Answer: 6.3\times 10^7m^3

Explanation:

Required amount of magnesium = 1.00\times 10^5 tons

Given : 1 ton = 2000 lb

1.00\times 10^5 tons=\frac{2000}{1}\times 1.00\times 10^5=2\times 10^8lb

1 lb = 453.592 g

2\times 10^8 lb=\frac{453.592 }{1}\times 2\times 10^8 lb=907.184\times 10^8g

Given : 1.4 g of magnesium is produced by 1000 g of sea water

907.184\times 10^8g of magnesium is produced by =\frac{1000}{1.4}\times 907.184\times 10^8g=6.5\times 10^{13} g of sea water

Density of sea water  = 1.025 g/ml

Volume of sea water =\frac{\text {mass of sea water}}{\text {density of sea water}}=\frac{6.5\times 10^{11}g}{ 1.025g/ml}=6.3\times 10^{13}ml

1 ml = 10^{-6}m^3

6.3\times 10^{13}ml=\frac{10^{-6}}{1}\times 6.3\times 10^{13}=6.3\times 10^7m^3

Volume of seawater, in cubic meters is 6.3\times 10^7

8 0
3 years ago
How many grams of sodium chloride, NaCl, result from the reaction shown in the following equation, FeCl3 + 3NaOH
LUCKY_DIMON [66]
<span>175.5g
</span>Cl(35.5) x 3 = 106.5
Na(23) x 3 = 69
106.5 + 69 = 175.5g

or

0g because sodium and chlorine are not chemically bonded.
7 0
4 years ago
Read 2 more answers
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