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ale4655 [162]
4 years ago
10

How can a permanent magnet be damaged?

Chemistry
1 answer:
34kurt4 years ago
3 0
<span>d. both the second and third choices above</span>
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Elaborate on how to classify elements and pure compounds.
tankabanditka [31]

Answer: C) Elements and pure compounds are homogeneous materials because they have a uniform composition throughout.

Explanation: Element is a pure substance which is composed of atoms of similar elements. Compound is a pure substance which is made from atoms of different elements combined together in a fixed ratio by mass.

Elements are compounds form homogeneous materials as they have uniform composition throughout and the components are evenly distributed throughout the material.

Mixtures are heterogeneous materials as they do not have uniform composition and the components are not evenly distributed throughout the material.

4 0
3 years ago
Read 2 more answers
In a second-order reaction (one that is second order in one reactant), cutting in half the concentration of that reactant will h
Keith_Richards [23]
rate 1 of the reaction = k [A]²

rate 2 of the reaction = k [A/2]² = k[A]² /4

We can see that rate of the reaction will be 4 times less.
8 0
4 years ago
8) Determine whether mixing each pair of the following results in a buffera. 100.0 mL of 0.10 M NH3 with 100.0 mL of 0.15 MNH4Cl
Marina86 [1]

Answer:

a. 100.0 mL of 0.10 M NH₃ with 100.0 mL of 0.15 M NH₄Cl.

c. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOH.

Explanation:

A buffer system is formed in 1 of 2 ways:

  • A weak acid and its conjugate base.
  • A weak base and its conjugate acid.

Determine whether mixing each pair of the following results in a buffer.

a. 100.0 mL of 0.10 M NH₃ with 100.0 mL of 0.15 M NH₄Cl.

YES. NH₃ is a weak base and NH₄⁺ (from NH₄Cl ) is its conjugate base.

b. 50.0 mL of 0.10 M HCl with 35.0 mL of 0.150 M NaOH.

NO. HCl is a strong acid and NaOH is a strong base.

c. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOH.

YES. HF is a weak acid and it reacts with NaOH to form NaF, which contains F⁻ (its conjugate base).

d. 175.0 mL of 0.10 M NH₃ with 150.0 mL of 0.12 M NaOH.

NO. Both are bases.

6 0
3 years ago
Some SO2 and O2 are mixed together in a flask at 1100 K in such a way ,that at the instant of mixing, their partial pressures ar
kakasveta [241]

Answer:

The answer is "\bold{0.525\ \ atm^{-1}}"

Explanation:

Given equation:

2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g)

Given value:

\Delta rH =-198.2 \ \ \frac{KJ}{mol}

Kp=1100 \ K

\Delta x = 2-(2+1)\\\\

     = 2-(2+1)\\\\= 2-(3)\\\\= -1

\left\begin{array}{cccc}I\ (atm)&1&0.5&0\\C\ (atm)&2x&-x&2x\\E\ (atm)     &1-2x&0.5-x&2x\end{array}\right

calculating the total pressure on equilibrium=  (1-2x)+(0.5-x)+2x \ atm\\\\

                                                                         = 1-2x+0.5-x+2x\\\\= 1+0.5-x\\\\=1.5-x\\

\therefore \\\\\to 1.50-x= 1.35 \\\\\to 1.50-1.35= x\\\\\to 0.15= x\\\\ \to  x= 0.15\\\\

calculating the pressure in  So_2:

= (1-2 \times 0.15)

= 1-0.30 \\\\ =0.70 \ atm

calculating the pressure in  O_2:

= (0.5- 0.15)\\\\= 0.35 \ atm \\

calculating the pressure in  So_3:

= (2 \times 0.15)\\\\= (.30) \ atm \\\\

Calculating the Kp at 1100 K:

= \frac{(Pressure(So_3))^2}{(Pressure(So_2))^2 \times Pressure(O_2)}\\\\= \frac{0.30^2}{0.70^2 \times 0.35}\\\\= \frac{0.30 \times 0.30 }{0.70\times 0.70 \times 0.35}\\\\= \frac{0.09 }{0.49\times 0.35} \\\\= \frac{0.09 }{0.1715} \\\\=  0.5247 \ \  or \ \  0.525 \ \ atm^{-1}  \\\\

4 0
3 years ago
What volume of 0.10 M NaOH is required to neutralize 30 mL of 0.10 M HCl?
coldgirl [10]
30 mL. The two compounds need to balance each other out.
5 0
3 years ago
Read 2 more answers
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