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kumpel [21]
2 years ago
9

A copper rod takes up 75 mL of space. With a density of 8.96 g/mL, what is the mass of the rod

Chemistry
1 answer:
pantera1 [17]2 years ago
5 0
5.73 is the mass of the rod
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Which of the following is a product in the reaction described by the word equation below?
Svet_ta [14]

Answer:

Please answer my question

6 0
2 years ago
5. A 5.00 g sample of an unknown substance was heated from 25.2 C to 55.1 degrees * C , and it required 133 to do so. Identify t
Leviafan [203]

Answer:

Aluminum

Explanation:

Given

T_1 = 25.2^oC

T_2 = 55.1^oC

m = 5.00g

\triangle Q= 133J

<em>See attachment for chart</em>

Required

Identify the unknown substance

To do this, we simply calculate the specific heat capacity from the given parameters using:

c = \frac{\triangle Q}{m\triangle T}

This gives:

c = \frac{\triangle Q}{m(T_2 - T_1)}

So, we have:

c = \frac{133J}{5.00g * (55.1C - 25.2C)}

c = \frac{133J}{5.00g * 29.9C}

c = \frac{133J}{149.5gC}

c = 0.89\ J/gC

From the attached chart, we have:

Al(s) = 0.89\ J/gC --- The specific heat capacity of Aluminum

<em>Hence, the unknown substance is Aluminum</em>

7 0
2 years ago
What type of information can one obtain by taking a mass spectrum of an organic molecule like dodecane?
Furkat [3]
Mass spectrum of Dodecane will give following information.

1 ) Molecular Peak or Parent Peak:
                                                       The Parent peak will appear at m/z = 170. The intensity of this peak will be very weak.

2) Fragments:
                      Usually the fragments of such long chain alkanes appear with spacing of 14 amu, Hence, the peaks in dodecane will be as follow,

         170 - 156 - 142 - 128 - 114 - 100 - 86 - 72 - 58 - 44 - 30 - 16

3) Base Peak:
                     Most probably the Base peak will appear at m/z = 57. This peak is due to the formation of tertiary butyl cation as the intensity mainly depends upon the stability of cation. So this cation might form due to rearrangment giving the intensity of 100%.
8 0
3 years ago
What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig
Mashutka [201]

Answer:

P_H =2.86

c=1.4\times 10^{-4}

Explanation:

first write the equilibrium equaion ,

C_3H_6O_3  ⇄ C_3H_5O_3^{-}  +H^{+}

assuming degree of dissociation \alpha =1/10;

and initial concentraion of C_3H_6O_3 =c;

At equlibrium ;

concentration of C_3H_6O_3 = c-c\alpha

[C_3H_5O_3^{-}  ]= c\alpha

[H^{+}] = c\alpha

K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}

\alpha is very small so 1-\alpha can be neglected

and equation is;

K_a = {c\alpha \times \alpha}

[H^{+}] = c\alpha = \frac{K_a}{\alpha}

P_H =- log[H^{+} ]

P_H =-logK_a + log\alpha

K_a =1.38\times10^{-4}

\alpha = \frac{1}{10}

P_H= 3.86-1

P_H =2.86

composiion ;

c=\frac{1}{\alpha} \times [H^{+}]

[H^{+}] =antilog(-P_H)

[H^{+} ] =0.0014

c=0.0014\times \frac{1}{10}

c=1.4\times 10^{-4}

6 0
3 years ago
In acidic solution, the sulfate ion can be used to react with a number of metal ions. One such reaction is SO42−(aq)+Sn2+(aq)→H2
allsm [11]

Answer:

The final balanced equation is :

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

Explanation:

SO_4^{2-}(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+Sn^{4+}(aq)

Balancing in acidic medium:

First we will determine the oxidation and reduction reaction from the givne reaction :

Oxidation:

Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)

Balance the charge by adding 2 electrons on product side:

Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)+2e^-....[1]

Reduction :

SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)

Balance O by adding water on required side:

SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)

Now, balance H by adding H^+ on the required side:

SO_4^{2-}(aq)+4H^+(aq)\rightarrow H_2SO_3(aq)+H_2O(l)

At last balance the charge by adding electrons on the side where positive charge is more:

SO_4^{2-}(aq)+4H^+(aq)+2e^-\rightarrow H_2SO_3(aq)+H_2O(l)..[2]

Adding [1] and [2]:

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

The final balanced equation is :

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

4 0
2 years ago
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