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yarga [219]
3 years ago
8

If an unknow metal has the mass of 40g and volume of 75ml, what is the density of the metal?

Chemistry
1 answer:
Snowcat [4.5K]3 years ago
6 0

Answer:

d = 0.53 g/mL

Explanation:

Given that,

Mass of an unknown metal = 40 g

The volume of the metal = 75 mL

We need to find the density of the metal. We know that,

Density = mass/ volume

d=\dfrac{40\ g}{75\ mL}\\\\d=0.53\ g/mL

So, the density of the metal is 0.53 g/mL.

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What did Ernest Rutherford s gold foil experiment demonstrate about atoms?
Ghella [55]
-Positively charged nucleus 
-Empty spaced
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5 0
3 years ago
Помогите пожалуйста или с первым вопросом или со вторым. Буду очень благодарна, т.к. это мой зачет... :С
Bess [88]
Hsiaqoanwbwiso iDisks
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3 years ago
For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia
Fittoniya [83]

Answer : The value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction:  Cu^{2+}+2e^-\rightarrow Cu

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu

To calculate the E^o_{(Cu^{2+}/Cu)} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}

Putting values in above equation, we get:

1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)

E^o_{(Cu^{2+}/Cu)}=0.34V

Hence, the value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

8 0
3 years ago
What is the percent composition of hydrogen in NH4HCO3?
Readme [11.4K]
To find the Percent Composition of an atom, you use this formula:
Mass of element in the compound you're studying on ( in this case it's 5 since there is 5 Hydrogens) over the mass of the compound (which is here 79), Multiplied by 100 since you want a percent. 
So we get: 
\frac{5}{79} * 100
So you get about: 
0.063 * 100
6.3

So, the percent composition of Hydrogen in NH4HCO3 is 6.3%

Hope this Helps! :D


7 0
3 years ago
Read 2 more answers
What is the volume in liters of 321 g of liquid with a density of 0.84 g/mL
Rasek [7]
Density is weight by volume.   

First.  If you divide the weight by density you can find the volume

Second you must convert the ML in to Liters.

\frac{321 \frac{g}{1}}{0.84 \frac{g}{mL}} = \frac{321(g)(mL)}{0.84g}=\frac{321mL}{0.84}=382.14mL

1L=1000ml

\frac{1L}{1000mL}

(382.14mL)( \frac{1L}{1000mL})= \frac{(382.14mL)(L)}{1000mL} =\frac{382.14L}{1000}=0.38214L

0.38214 Liters.

4 0
3 years ago
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