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UkoKoshka [18]
3 years ago
14

True or false: an amplifier can adjust the amplitude

Chemistry
1 answer:
DENIUS [597]3 years ago
6 0

Answer:

<h3>ᴛʀᴜᴇ..</h3>

...............

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Diazomethane, CH2N2, is used in the organic chemistry laboratory despite its danger because it produces very high yields and is
dexar [7]

Answer:

See picture below

Explanation:

You are not providing the starting compound. However, I found a similar question so, I will draw the product of this compound, and then, you follow the same procedure.

As the problem states, the diazomethane is often used in reaction with carboxilic acid. This kind of reaction will do an esterification reaction, and the final product will always result in an esther.

So, with the example I give you here, all you have to do is replace the OH in the carboxilic group of your initial compound, for the OCH₃ group.

See picture below.

Hope this helps

5 0
3 years ago
P4 +6Br2 --&gt; 4PBr3
grin007 [14]

Answer:

1 mole of P₄ consumed:4 moles of PBr₃ produced

Explanation:

Look at the coefficients in the balanced chemical equation to determine the ratios. For every 1 mole of P₄ consumed (P₄ has a coefficient of 1), 4 moles of PBr₃ would be produced (PBr₃ has a coefficient of 4).

6 0
3 years ago
Read 2 more answers
Suppose 40.8g of copper(II) acetate is dissolved in 200.mL of a 0.70 M aqueous solution of sodium chromate.
Contact [7]

Answer:

0.42 M

Explanation:

The reaction that takes place is:

  • Cu(CH₃COO)₂ + Na₂CrO₄ → Cu(CrO₄) + 2Na(CH₃COO)

First we <u>calculate the moles of Na₂CrO₄</u>, using the <em>given volume and concentration</em>:

(200 mL = 0.200L)

  • 0.70 M * 0.200 L = 0.14 moles Na₂CrO₄

Now we <u>calculate the moles of Cu(CH₃COO)₂</u>, using its <em>molar mass</em>:

  • 40.8 g ÷ 181.63 g/mol = 0.224 mol Cu(CH₃COO)₂

Because the molar ratio of Cu(CH₃COO)₂ and Na₂CrO₄ is 1:1, we can directly <u>substract the reacting moles of Na₂CrO₄ from the added moles of Cu(CH₃COO)₂</u>:

  • 0.224 mol - 0.14 mol = 0.085 mol

Finally we <u>calculate the resulting molarity</u> of Cu⁺², from the <em>excess </em>cations remaining:

  • 0.085 mol / 0.200 L = 0.42 M

4 0
3 years ago
A 21.5 g sample of nickel was treated with excess silver nitrate solution to produce silver metal and nickel(II) nitrate. The re
Arte-miy333 [17]

The mass of nickel is 36.5g metal.

Ni+2AgNO3 → Ni(NO3)2+2Ag

21.5g Ni / 58.69g = 0.366mol Ni

Ag = 107.87

N i= 58.69

But, 2molAg per 1 mol Ni, so, change in mass is more like...

Twice as many moles Ag as Ni, and then the mass relationship is 107.87g Ag / 58.69, but due to the 2Ag: 1Ni mol ratio, the change in mass is like:

107.87(2) / 58.69 = 3.68 times increase

So, the final mass of Ag is 3.68x, and the change in mass of Ni is x

36.5=(21.5-x)+(3.68x)

Solve for x

x=5.597

3.68x=g Ag

3.68(5.597)=20.59

21.5 - x = g Ni after reaction

21.5 - 5.597 = 15.903g Ni

20.59g Ag + 15.903g Ni = 36.5g metal

The mass of nickel is 36.5g metal.

Learn more about reactions here: brainly.com/question/16416932

#SPJ4

7 0
2 years ago
4. Solve the following heat flow problem, being sure to show all your work (you may either type your
Viktor [21]

Answer:

0.70 J/g.°C

Explanation:

Step 1: Given data

  • Mass of graphite (m): 402 g
  • Heat absorbed (Q): 1136 J
  • Initial temperature: 26°C
  • Final temperature: 30 °C
  • Specific heat of graphite (c): ?

Step 2: Calculate the specific heat of graphite

We will use the following expression.

Q = c × m × ΔT

c = Q / m × ΔT

c = 1136 J / 402 g × (30°C - 26°C)

c = 0.70 J/g.°C

5 0
4 years ago
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