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vivado [14]
3 years ago
14

Is it general principal that more abundant isotopes exist in

Chemistry
1 answer:
Nataly_w [17]3 years ago
5 0

sjnsjshdjahshabbxbabsbasjsjjsjs

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Gold is alloyed (mixed) with other metals to increase its hardness in making jewelry.
KiRa [710]

Answer:

A) 54.04%

B) 13-karat

Explanation:

A) From the problem we have

<em>1)</em> Mg + Ms = 9.40 g

<em>2)</em> Vg + Vs = 0.675 cm³

Where M stands for mass, V stands for volume, and g and s stand for gold and silver respectively.

We can rewrite the first equation using the density values:

<em>3)</em> Vg * 19.3 g/cm³ + Vs * 10.5 g/cm³ = 9.40

So now we have<em> a system of two equations</em> (2 and 3) <em>with two unknowns</em>:

We <u>express Vg in terms of Vs</u>:

  • Vg + Vs = 0.675 cm³
  • Vg = 0.675 - Vs

We <u>replace the value of Vg in equation 3</u>:

  • Vg * 19.3 + Vs * 10.5 = 9.40
  • (0.675-Vs) * 19.3 + Vs * 10.5 = 9.40
  • 13.0275 - 19.3Vs + 10.5Vs = 9.40
  • -8.8 Vs + 13.0275 = 9.40
  • <u>Vs = 0.412 cm³</u>

Now we <u>calculate Vg</u>:

  • Vg + Vs = 0.675 cm³
  • Vg + 0.412 cm³ = 0.675 cm³
  • Vg = 0.263 cm³

We <u>calculate Mg from Vg</u>:

  • 0.263 cm³ * 19.3 g/cm³ = 5.08 g

We calculate the mass percentage of gold:

  • 5.08 / 9.40 * 100% = 54.04%

B)

We multiply 24 by the percentage fraction:

  • 24 * 54.04/100 = 12.97-karat ≅ 13-karat
4 0
3 years ago
Tính axit giảm dần của các chất C6H5 OH (1) ,p – CH3 OC6H4 OH (2), p-NO2C6H4OH (3) pCH3COC6H4OH (4), p-CH3C6H4 OH (5) là:
nadya68 [22]

Answer:

di ko po Alam sorry

Explanation:

sorry sorry sorry sorry sorry

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3 years ago
Which of the following examples describes an increase in potential energy? Question 1 options: a basketball falling through a ne
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Paint being carried up a ladder
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3 years ago
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Copper was the first metal to be produced from its ore because it is the easiest to smelt, that is, to refine by heating in the
Ganezh [65]

Answer:

57.48%

Explanation:

Calculate the mass of 1 mole of malachite:

MM Cu = 63.55

MM O = 16.00

MM H = 1.01

MM C = 12.01

(Cu_{2}(OH)_{2}CO_{3})

A mole of malachite has:

2 moles of Cu

5 moles of O

2 moles of H

1 mole of C

MW Malachite = 2*MM(CU) + 5*MM(O) + 2*MM(H) + 1 *MM(C)

MW Malachite = 2*63.55 + 5*16.00 + 2*1.01 + 1*12.01

MW Malachite = 221.13

Mass of Cu in a mole of Malachite = 2*MM(CU) = 127.1

Now divide the mass of Cu by the mass of Malachite

%Cu = \frac{127.1}{221.13} =0.5748=57.48%

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Drug designers are usually
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