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Andrews [41]
3 years ago
7

What is the voltage this voltmeter is measuring? Be sure your answer has a reasonable number of decimal place

Chemistry
1 answer:
Mnenie [13.5K]3 years ago
8 0
Hi! I would say the voltage can be measured at 0.083 volts
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Why is baking soda an ionic bond
Kaylis [27]
Because it is made up of sodium, a metal, and carbon, a nonmetal.

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3 years ago
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Given the following reaction: NH4SH (s) <--> NH3 (g) + H2S (g) If we start
almond37 [142]

Answer:

D. 0.3 M

Explanation:

                                              NH4SH (s)      <-->            NH3 (g) + H2S (g)

Initial concentration              0.085mol/0.25L             0                 0

Change in concentration     -0.2M                               +0.2 M        +0.2M

Equilibrium               0.035mol/0.25 L=0.14M             0.2M           0.2M

concentration

Change in concentration (NH4SH) = (0.085-0.035)mol/0.25L =0.2M

K = [NH3]*[H2S]/[NH4SH] = 0.2M*0.2M/0.14M ≈ 0.29 M ≈ 0.3M

4 0
3 years ago
0.000068 g Express your answer as an integer.
ivann1987 [24]
6.8x10^5? Scientific notation
5 0
3 years ago
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Which of these can be described as a system of stars, gases, dust, and other matter that orbits a common center of gravity?
velikii [3]
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8 0
2 years ago
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A 2.00L flask was filled with 4.00 mol of HI at a certain temperature and given sufficient time to react. At equilibrium the con
inn [45]

Answer:

The equilibrium concentration of I₂ is 0.400 M  and HI is 1.20 M, the Keq will be 0.112.

Explanation:

Based on the given information, the equilibrium reaction will be,  

2HI (g) ⇔ H₂ (g) + I₂ (g)

It is given that 4.00 mol of HI was filled in a flask of 2.00 L, thus, the concentration of HI will be,  

= 4.00 mol/2.00 L

= 2.00 mol/L

Based on the reaction, the initial concentration of 2HI is 2.00, H₂ is 0 and I₂ is O. The change in the concentration of 2HI is -x, H₂ is x and I₂ is x. The equilibrium concentration of 2HI will be 0.200-x, H₂ is x and I₂ is x.  

It is given that at equilibrium, the concentration of H₂ or x is 0.400 M.  

Now the equilibrium concentration of HI will be,  

= 2.00 -2x  

= 2.00 - 2 × 0.400

= 1.20 M

The equilibrium concentration of I₂ will be,  

I₂ = x  

= 0.400 M

The equilibrium constant (Keq) will be,  

Keq = [H₂] [I₂] / [HI]²

= (0.400) (0.400) / (1.20)²

= 0.112

Thus, the Keq of the reaction will be 0.112.  

5 0
3 years ago
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