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timofeeve [1]
2 years ago
9

if the density of copper is 8.9g/cm^3 and the Volume of the sample is 23.4 mL what would be the mass? ​

Chemistry
1 answer:
Black_prince [1.1K]2 years ago
8 0

Answer:

The mass of copper = 208.26 grams.

Explanation:

Density = mass / volume

8.9 = mass / 23.4

mass = 8.9 * 23.4

= 208.26 grams.

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Answer:

'Wave' is a common term for a number of different ways in which energy is transferred: In electromagnetic waves, energy is transferred through vibrations of electric and magnetic fields. In sound waves, energy is transferred through vibration of air particles or particles of a solid through which the sound travels.

Explanation:

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when washing soda is mixed with lemon juice, bubbles are formed with the evolution of gas. what type of change is it. ​
aleksandr82 [10.1K]

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it is a chemical change

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Measurements show that unknown compound X has the following composition: element mass % carbon 41.0% hydrogen 4.58% oxygen 54.6%
anastassius [24]

Answer:

CHO

Explanation:

Carbon = 41%,  Hydrogen = 4.58%, oxygen = 54.6%

Step 1:

Divide through by their respective relative atomic masses

41/ 12,         4.58/1,         54.6/16

3.41              4.58            3.41

Step 2:

Divide by the lowest ratio:

3.41/3.41,      4.58/3.41,     3.41/3.41

1,                    1,                  1

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8 0
3 years ago
A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after
choli [55]

Answer:

9.25

Explanation:

Let first find the moles of NH_4Cl and NaOH

number of moles of NH_4Cl = 0.40  mol/L × 200 ×  10⁻³L

= 0.08 mole

number of moles of NaOH = 0.80  mol/L × 50 ×  10⁻³L

= 0.04 mole

The equation for the reaction is expressed as:

NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

The ICE Table is shown below as follows:

                            NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

Initial (M)              0.08            0.04                            0

Change (M)         - 0.04          -0.04                          + 0.04

Equilibrium (M)      0.04             0                                0.04

K_a*K_b = 10^{-14} \ at \ 25^0C

K_a = \frac{10^{-14}}{K_b}

K_a = \frac{10^{-14}}{1.76*10^{-5}}

K_a= 5.68*10^{10}

pK_a = - log \ (K_a)

pK_a = - log \ (5.68*10^{-10})

pK_a = 9.25

pH = pKa + \ log (\frac{HB}{HA} )   for buffer solutions

pH = pKa + \ log (\frac{moles \ of \ base }{ moles\ of \ acid} ) since they are in the same solution

pH = 9.25 + \ log (\frac{0.04 }{ 0.04} )

pH = 9.25

8 0
3 years ago
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