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mylen [45]
3 years ago
7

A Gold ring is a what? Element Compound Colloid Alloy

Chemistry
1 answer:
Korvikt [17]3 years ago
4 0

Answer: Alloy

Explanation: gold is an alloy, or mixture of metals

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What is the different between a graph representing data that is directly proportional and a graph that is inversely proportional
sesenic [268]
A graph depicting a direct relation is a straight line and usually has positive slope. An inverse relation is a curve, typically concave up with negative slope.
7 0
3 years ago
A naturally occurring oil co-distills with water to produce an oil/water distillate that is 20% oil by weight. If the molecular
Gelneren [198K]

Answer:

Explanation:

Partial pressure of oil = mole fraction of oil x total pressure

mole fraction of oil = mole of oil / mole of water + mole of oil

= mole of oil = mass of oil / molecular weight of oil

= 20 / 100 = .2

mole of water = 80 / 18

= 4.444

mole fraction of oil =  .2 / .2 + 4.444

= .2 / 4.644

Partial pressure of oil = mole fraction of oil x total pressure

= (.2 / 4.644 ) x 760 mm

= 32.73 mm Hg .

3 0
3 years ago
A solution is made by mixing 37.g of thiophene C4H4S and 72.g of heptane C7H16. Calculate the mole fraction of thiophene in this
guajiro [1.7K]

Answer:

0.38

Explanation:

Molar mass of thiophene= 84g/mol

Mass of thiophene = 37g

Number of moles= 37/84= 0.44 moles

Molar mass of heptane= 100 g/mol

Mass of heptane = 72g

Number of moles = 72/100= 0.72 moles

Total number of moles= 0.44 + 0.72= 1.16 moles

mole fraction of thiophene = 0.44/1.16= 0.38

4 0
3 years ago
The number of kilograms of water in a human body varies directly as the mass of the body. upper a 93​-kg person contains 62 kg o
Bad White [126]

Answer;

56 kg of water


Solution;

Weight = 62 kg of water in 93 kg of a person

W=km

62=93k;

k=2/3

W=(2/3)84

=56 kg

Therefore; a 84-kg person will have 56 kg of water.

4 0
3 years ago
How many grams of sulfuric acid is needed to neutralize 380 ml of solution with pH = 8.94
erma4kov [3.2K]

Answer : The mass of sulfuric acid needed is 16.23\times 10^{-5}g.

Solution : Given,

pH = 8.94

Volume of solution = 380 ml = 380\times 10^{-3}      (1ml=10^{-3}L)

Molar mass of sulfuric acid = 98.079 g/mole

As we know,

pH+pOH=14\\pOH=14-8.94=5.06

pOH=-log[OH^-]

5.06=-log[OH^-]

[OH^-]=0.00000871=8.71\times 10^{-6}mole/L

Now we have to calculate the moles of OH^-.

Formula used : Moles=Concentration\times Volume

\text{ Moles of }[OH^-]=\text{ Concentration of }[OH^-]\times Volume\\\text{ Moles of }[OH^-]=(8.71\times 10^{-6}mole/L)\times (380\times 10^{-3}L)=3309.8\times 10^{-9}moles

For neutralization, equal number of moles of H^+ ions will neutralize same number of OH^- ions.

\text{ Moles of }[OH^-]=\text{ Moles of }[H^+]=3309.8\times 10^{-9}moles

As, H_2SO_4\rightarrow 2H^++SO^{2-}_4

From this reaction, we conclude that

2 moles of H^+ ion is given by the 1 mole of H_2SO_4

3309.8\times 10^{-9} moles of H^+ ion is given by \frac{3309.8\times 10^{-9}}{2}=1654.9\times 10^{-9} moles of H_2SO_4

Now we have to calculate the mass of sulfuric acid.

Mass of sulfuric acid = Moles of H_2SO_4 × Molar mass of sulfuric acid

Mass of sulfuric acid = (1654.9\times 10^{-9}moles)\times (98.079g/mole)=162310.94\times 10^{-9}=16.23\times 10^{-5}g

Therefore, the mass of sulfuric acid needed is 16.23\times 10^{-5}g.

3 0
3 years ago
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