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saul85 [17]
3 years ago
14

The ability of nitrogen to attract electrons is

Chemistry
1 answer:
Advocard [28]3 years ago
8 0

Answer:

Oxygen

Explanation:

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1a. A researcher accidentally spilled some sand into his beaker containing crystalline compound X. Knowing what he did about mel
Sunny_sXe [5.5K]

Answer:

See explanation

Explanation:

Many organic compounds have low melting points. This is due to the fact that many of these compounds are non polar.

However, compound X is slightly polar but still has a melting point which is far less than that of sand composed of a high melting point inorganic material.

Since sand has a much higher melting point compared to compound X, the researcher need not be worried that sand was spilled into his beaker.

7 0
2 years ago
A chemist designs a galvanic cell that uses these two half-reactions: half-reaction standard reduction potential (s)(aq)(aq)(l)
miv72 [106K]

Answer:

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)  

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻        

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

E°cell = 1.10 V

Explanation:

<em>The half-reactions are missing, but I will propose some to show you the general procedure and then you can apply it to your equations.</em>

<em>Suppose we have the following half-reactions.</em>

<em>Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)   E°red = 0.34 V</em>

<em>Zn²⁺(⁺aq) + 2 e⁻ → Zn(s)    E°red = -0.76 V</em>

<em />

To identify how to make a spontaneous cell, we need to consider the standard reduction potentials (E°red). The half-reaction with the higher E°red will occur as a reduction (in the cathode), whereas the one with the lower E°red will occur as an oxidation (in the anode).

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)   E°red = 0.34 V

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻        E°red = -0.76 V

To get the overall equation we add both half-reactions.

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an

E°cell = 0.34 V - (-0.76 V) = 1.10 V

Since E°cell > 0, the reaction is spontaneous.

5 0
3 years ago
Please help with #2 and #3
Vaselesa [24]

Answer:

2. V_2=17L

3. V=82.9L

Explanation:

Hello there!

2. In this case, we can evidence the problem by which volume and temperature are involved, so the Charles' law is applied to:

\frac{V_2}{T_2}=\frac{V_1}{T_1}

Thus, considering the temperatures in kelvins and solving for the final volume, V2, we obtain:

V_2=\frac{V_1T_2}{T_1}

Therefore, we plug in the given data to obtain:

V_2=\frac{18.2L(22+273)K}{(45+273)K} \\\\V_2=17L

3. In this case, it is possible to realize that the 3.7 moles of neon gas are at 273 K and 1 atm according to the STP conditions; in such a way, considering the ideal gas law (PV=nRT), we can solve for the volume as shown below:

V=\frac{nRT}{P}

Therefore, we plug in the data to obtain:

V=\frac{3.7mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm}\\\\V=82.9L

Best regards!

6 0
3 years ago
PLEASE HELP DUE SOON&gt; WILL MARK BRAINLIEST 50 POINTS
Rainbow [258]
Methane CH4
ethane C2H6
propane C3H8
Butane C4H10
pentane C5H12
8 0
3 years ago
Read 2 more answers
Zn + I2 ---&gt; Znl2<br> Determine the theoretical yield of the product if 2g of Zn is used
egoroff_w [7]

Answer:

Mass = 9.58 g

Explanation:

Given data:

Mass of Zn = 2g

Theoretical yield of ZnI₂ = ?

Solution:

Chemical equation:

Zn + I₂       →     ZnI₂

Number of moles of Zn:

Number of moles = mass/molar mass

Number of moles = 2g / 65.38 g/mol

Number of moles = 0.03 mol

Now we will compare the moles of Zn and ZnI₂.

                   Zn           :          ZnI₂

                    1             :           1

                  0.03        :         0.03

Mass of ZnI₂:

Mass = number of moles × molar mass

Mass = 0.03 mol × 319.22 g/mol

Mass = 9.58 g

4 0
3 years ago
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