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pogonyaev
3 years ago
6

If the pressure on a gas is decreased by one-half, how large will the volume change be?

Chemistry
1 answer:
Jet001 [13]3 years ago
3 0

Answer:

It will double in size

Explanation:

i hope its correct

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How does the reaction of hydrogen and oxygen to produce energy in a fuel cell differ from their interaction during the direct co
Nikolay [14]
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7 0
3 years ago
If a sample of pure cadmium has a mass of 18.2 grams, how many moles of cadmium are in the sample?
Mars2501 [29]
Atomic mass cadmium = 112.41 amu

1 mole Cd ------------ 112.41 g
?? moles Cd --------- 18.2 g

18.2 x 1 / 112.41 => 0.161 moles of Cd

hope this helps!
3 0
3 years ago
Important (question on k12 test)
monitta

Answer:

A new Dana system of classification contains 78 different classes of minerals based on composition and then further classified by type and group. To be considered a mineral, a substance must be an inorganic, naturally formed solid, with a specific chemical formula and a fixed internal structure. To test whether something is a mineral, there are several identification tests to which the substance is subjected, including its resistance to scratching, its density in comparison to water, its color, the degree of light it reflects, the color of the powdered mineral, its breakage pattern and its crystalline form.

Explanation:

4 0
3 years ago
Read 2 more answers
What is the concentration of x2??? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5??10???6 and ka2=1.2??10???11
lutik1710 [3]
The first dissociation for H2X:
                        H2X +H2O ↔ HX + H3O
initial                0.15                     0      0
change             -X                     +X      +X
at equlibrium 0.15-X                  X        X
because Ka1 is small we can assume neglect x in H2X concentration
     Ka1      = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15) 
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
        HX + H2O↔ X^2 + H3O
    8.2x10^-4          Y         8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2       = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m


4 0
3 years ago
Hydrofluoric acid (HF) can be prepared according to the following equation:
Artemon [7]

Given the balanced equation:

( Reaction type : double replacement)

CaF2 + H2SO4 → CaSO4 + 2HFI

We can determine the number of grams prepared from the quantity of 75.0 H2SO4, and 63.0g of CaF2 by converting these grams to moles per substance.

This can be done by evaluating the atomic mass of each element of the substance, and totaling it to find the molecular mass.

For H2SO4 or hydrogen sulfate it's molecular mass is the sum of the quantity of atomic mass per element. H×2 + S×1 + O×4 = ≈1.01×2 + ≈32.06×1 + ≈16×4 = 2.02 + 32.06 + 64 = 98.08 u (Dalton's or Da) or g / mol.

For CaF2 or calcium fluoride, it's molecular mass adds 1 atomic mass of calcium and 2 atomic masses of fluoride due to the number of atoms.

Ca×1 + F×2 = ≈40.07×1 + ≈19×2 = 40.08 + 38 = 78.07 u (Da or Dalton's) or g / mol.

3 0
3 years ago
Read 2 more answers
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