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poizon [28]
3 years ago
14

Which statment correctly decribes a law?

Chemistry
1 answer:
xeze [42]3 years ago
4 0
B) is a factual statement
Laws are based on observations and experiments and have been tested many, many times to show no error
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What are valence electrons? how many of a magnesium atom’s 12 electrons are valence electrons?
wlad13 [49]

It has 2 valence electrons.

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3 years ago
A certain ore is 37.3% nickel by mass how many kilograms of this ore would you need to dig up to have 10.0g of nickel
jeka57 [31]
If the grade of the ore is 37.3% nickel, then the unknown quantity to get 10 grams of nickel is 0.373 x = 10 grams or x = 10/0.373=26.8 grams or 0.0268 kg needed to dig up to recover the 10 grams of nickel. At this grade of ore, 1 kilogram would yield 373 grams of nickel. 
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Is LiOH soluble or insoluble?
riadik2000 [5.3K]
LiOH is soluble. Na2CO3 is soluble. Cu(OH)2 is insoluble. 
6 0
3 years ago
Determine the mass of a gold bar that has a density of 19.3 g/cm3 and is 4.72 cm high by 8.21 cm long by 3.98 cm deep.
TEA [102]
The density of the gold is 19.3 grams/cc so each cc weighs 19.3grams. Now we can obtain the volume of gold from the given dimensions ie 4.72x8.21x3.98= 154.23 cc. So for the answer, just multiply the volume or 154.23 x 19.3= 2976.6 grams is the answer.
5 0
3 years ago
A sample of gas in a balloon has an initial temperature of 20. ∘C and a volume of 1.92×103 L . If the temperature changes to 68
morpeh [17]

Answer:

\boxed{2.23 \times 10^{3} \text{ L}}

Explanation:

The pressure is constant, so we can use Charles' Law.

\dfrac{ V_{1} }{T_{1}} = \dfrac{ V_{2} }{T_{2}}

Data:

V₁ = 1.92 × 10³ L; T₁ = 20 °C  

V₂ = ?;                 T₂ = 68 °C

Calculations:

(a) Convert temperatures to kelvins

T₁ = (20 + 273.15) K = 293.15 K

T₂ = (68 + 273.15) K = 341.15 K

(b) Calculate the volume

\dfrac{ 1.92 \times 10^{3}}{293.15} = \dfrac{ V_{2}}{341.15}\\\\6.550 = \dfrac{ V_{2}}{341.15}\\\\V_{2} = 6.550 \times 341.15 = 2.23 \times 10^{3} \text{ L}

The new volume of the gas is \boxed{2.23 \times 10^{3} \text{ L}}.

6 0
3 years ago
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