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pychu [463]
3 years ago
10

Why is the mass of an iron atom different from the mass of a copper atom?

Chemistry
1 answer:
Jobisdone [24]3 years ago
6 0
They have different number of Neutrons and protons, so their masses are different

Hope this helps!
You might be interested in
What mass of Cu(s) is electroplated by running 24.5A of current through a Cu2+(aq)solution for 4.00 h?Express your answer to thr
Masja [62]

Answer: 116 g of copper

Explanation:

Q=I\times t

where Q= quantity of electricity in coloumbs

I = current in amperes = 24.5A

t= time in seconds =  4.00 hr = 4.00\times 3600s=14400s  (1hr=3600s)

Q=24.5A\times 14400s=352800C

Cu^{2+}+2e^-\rightarrow Cu

2\times 96500C=193000C  of electricity deposits 63.5 g of copper.

352800 C of electricity deposits = \frac{63.5}{193000}\times 352800=116g of copper.

Thus 116 g of Cu(s) is electroplated by running 24.5A of current

Thus  remaining in solution = (0.1-0.003)=0.097moles

8 0
3 years ago
Which of the following decomposition reactions is/are correct?
Valentin [98]
<span>The generalized reaction for chemical decomposition is: AB → A + B

NaOH is sodium hydroxide. When sodium and water is combined it makes sodium hydroxide and hydrogen

When sodium hydroxide decomposes under thermal decomposition, it breaks down into sodium oxide and water.

Thus, </span><span>C) 2NaOH Na2O + H2O</span>
4 0
3 years ago
What's the empirical and molecular weight of C14H22N4O8? What's the molecular formula of C14H22N4O8?
Inessa05 [86]

374u

187u

C₁₄H₂₂N₄O₈

Explanation:

To find the molecular weight of the compound C₁₄H₂₂N₄O₈ we simply sum that atomic masses of the given elements in the compound.

 The empirical weight is determined by using the simplest ratio of the elements involved in the compound;

Molecular weight of C₁₄H₂₂N₄O₈;

atomic mass of C = 12g/mol

                           H = 1g/mol

                            N = 14g/mol

                            O = 16g/mol

 Molecular weight = 14(12) + 22(1) + 4(14) + 8(16)

                                = 168 + 22 + 56 + 128

                                 = 374u

Empirical weight:

  Empirical formula:

                        C₁₄    H₂₂      N₄     O₈

                         14  :    22 :    4  :     8

   divide by 2:

                          7   :    11    :    2  :    4

     empirical formula  C₇H₁₁N₂O₄

     empirical weight = \frac{molecular weight}{2} = \frac{374}{2} = 187u

The molecular formula is the actual combination of atoms in a compound. so the molecular formula of the compound is C₁₄H₂₂N₄O₈

learn more:

Molecular mass brainly.com/question/5546238

#learnwithBrainly

8 0
3 years ago
The normal boiling point of a liquid is 282 °C. At what temperature (in
ElenaW [278]

Answer:

The temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C

Explanation:

Here we make use of the Clausius-Clapeyron equation;

ln\left (\frac{p_{2}}{p_{1}}  \right )=-\frac{\Delta H_{vap}}{R}\cdot \left (\frac{1}{T_{2}}-\frac{1}{T_{1}}  \right )

Where:

P₁ = 1 atm =The substance vapor pressure at temperature T₁ = 282°C = 555.15 K

P₂ = 0.2 atm = The substance vapor pressure at temperature T₂

\Delta H_{vap} = The heat of vaporization = 28.5 kJ/mol

R = The universal gas constant = 8.314 J/K·mol

Plugging in the above values in the Clausius-Clapeyron equation, we have;

ln\left (\frac{0.2}{1}  \right )=-\frac{28.5 \times 10^3}{8.3145}\cdot \left (\frac{1}{T_{2}}-\frac{1}{555.15}  \right )

\therefore T_2 = \frac{-3427.95}{ln(0.2)-6.175}

T₂ = 440.37 K

To convert to Celsius degree temperature, we subtract 273.15 as follows

T₂ in °C = 440.37 - 273.15 = 167.22 °C

Therefore, the temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C.

5 0
3 years ago
Which reaction describes an alpha emission?
Tcecarenko [31]

88226Ra→ 86222Rn + 24He is the reaction that describes an alpha emission because radiations are released.

<h3>Describes an alpha emission?</h3>

Alpha radiation occurs when the nucleus of an atom becomes unstable and alpha particles are released in order to restore stability. Alpha decay occurs in elements that have high atomic numbers, such as uranium, radium, and thorium etc.

So we can conclude that 88226Ra→ 86222Rn + 24He is the reaction that describes an alpha emission because radiations are released.

Learn more about radiation here: brainly.com/question/893656

#SJ1

7 0
2 years ago
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