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DedPeter [7]
2 years ago
11

Which of these statements about alloys and intermetallic compounds is false? Which of these statements about alloys and intermet

allic compounds is false? Intermetallics are compounds of two or more metals that have a definite composition and are not considered alloys. Bronze is an example of an alloy. "Alloy" is just another word for "a chemical compound of fixed composition that is made of two or more metals." Alloys can be formed even if the atoms that comprise them are rather different in size. If you mix two metals together and, at the atomic level, they separate into two or more different compositional phases, you have created a heterogeneous alloy.
Chemistry
1 answer:
Black_prince [1.1K]2 years ago
3 0

The statement that is an alloy is just another word for a chemical compound of fixed composition that is made of two or more metals is false.  

An alloy refers to the mixture of the elements in which the parent element is the metal. An alloy is produced by the amalgamation of any element, that is, metal or non-metal to better the feature of the parent metal. The whole components in the alloy are combined to produce a mixture of fixed composition, but not a chemical compound, and it is not essential that the elements involved are metals. Thus, the mentioned statement is false.  


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Atorvastatin is sold under the trade name Lipitor and is used for lowering cholesterol. Annual global sales of this compound exc
kotegsom [21]

Answer:

Atorvastatin has two chiral centers. The question doesn't include the box where have to answer but I can show you in an image where are located and their configuration.

Explanation:

The first image shows the chemical structure of atorvastatin and their chiral centers identified as 1 and 2 respectively.

The second image shows the Fischer projections corresponding to every chiral carbon 1 and 2. I wrote R so suggest that there are more carbon atoms forward but not only corresponds to carbon atoms.

You can see that the chiral carbon 1 has R configuration due to the direction from the main substituent to the second follow the clockwise.

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2 years ago
Is utensils a substance homogeneous mixture or heterogeneous mixture
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7 0
2 years ago
Picric acid has been used in the leather industry and in etching copper. However, its laboratory use has been restricted because
olchik [2.2K]

Answer:

0.3023 M

Explanation:

Let Picric acid = H_{picric}

So,  H_{picric}     +       H_2}O          ⇄      H_3}O^+     +     Picric^-

The ICE table can be given as:

                          H_{picric}     +       H_2}O          ⇄      H_3}O^+     +     Picric^-

Initial:                0.52                                               0                  0

Change:             - x                                                 + x                 + x

Equilibrium:      0.52 - x                                        + x                 + x

Given that;

acid dissociation constant  (K_a) = 0.42

K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}

0.42 = \frac{[x][x]}{0.52-x}}

0.42 = \frac{[x]^2}{0.52-x}}

0.42(0.52-x) = x²

0.2184 - 0.42x = x²

x²  + 0.42x - 0.2184 = 0                   -------------------- (quadratic equation)

Using the quadratic formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}    ;     ( where +/-  represent ± )

= \frac{-0.42+/-\sqrt{(0.42)^2-4(1)(-0.2184)} }{2*1}

= \frac{-0.42+/-\sqrt {0.1764+0.8736} }{2}

= \frac{-0.42+\sqrt {1.0496} }{2}     OR   \frac{-0.42-\sqrt {1.0496} }{2}

= \frac{-0.42+1.0245}{2}       OR    \frac{-0.42-1.0245}{2}

= \frac{0.6045}{2}                 OR    -\frac{1.4445}{2}

= 0.30225          OR     - 0.72225

So, we go by the +ve integer that says:

x =  0.30225

x = [ H_3}O^+ ] = [   Picric^- ] =  0.3023  M

∴  the value of  [H3O+] for an 0.52 M solution of picric acid  = 0.3023 M     (to 4 decimal places).

6 0
2 years ago
During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.
Monica [59]

Answer:

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

  • CH₄ + 2 O₂ → CO₂ + 2 H₂O

it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) /  (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) /  (32.00 g/mol) = 0.2812 mol.

  • it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of  CH₄ needs → 2 mol of O₂

???  mol of  CH₄  needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over *  molar mass

                                    = 0.0469 mol * 16 g/mol = 0.7504 g

So, the right choice is 0.74 grams of methane

3 0
2 years ago
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