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pishuonlain [190]
2 years ago
14

13500 mL convert to milliliters of mercury

Chemistry
1 answer:
larisa86 [58]2 years ago
8 0

Answer:approximately 5.627 × 1019 milliliters

Explanation:

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Fofino [41]

Answer:

when u do the dishes and when you take a shower i thank if you put thos in a full sentence they will be good answer

Explanation:

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3 years ago
How much of a chemical should you take?
barxatty [35]

Answer:

No more than 0.1 mL of hydrochloricton acid

5 0
2 years ago
Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.
Ad libitum [116K]

Explanation:

The given data is as follows.

    Concentration = 0.1 mol/dm^{3}

                             = 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

                             = 6.022 \times 10^{25} ions/m^{3}

               T = 30^{o}C = (30 + 273) K = 303 K

Formula for electric double layer thickness (\lambda_{D}) is as follows.

            \lambda_{D} = \frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}

where, n^{o} = concentration = 6.022 \times 10^{25} ions/m^{3}

Hence, putting the given values into the above equation as follows.

                 \lambda_{D} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}                    

                          = \sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}  

                         = 9.669 \times 10^{-10} m

or,                     = 9.7 A^{o}

                          = 1 nm (approx)

Also, it is known that \lambda_{D} = \sqrt \frac{1}{n^{o}}

Hence, we can conclude that addition of 0.1 mol/dm^{3} of KCl in 0.1 mol/dm^{3} of NaBr "\lambda_{D}" will decrease but not significantly.

7 0
3 years ago
What is the theoretical yield of NaBr
dolphi86 [110]

The theoretical yield of NaBr given that 2.36 moles of FeBr₃ reacts is 7.08 moles

<h3>Balanced equation </h3>

2FeBr₃ + 3Na₂S → Fе₂S₃ + 6NaBr

From the balanced equation above,

2 moles FeBr₃ reacted to produce 6 moles of NaBr

<h3>How to determine the theoretical yield of NaBr</h3>

From the balanced equation above,

2 moles FeBr₃ reacted to produce 6 moles of NaBr

Therefore,

2.36 moles FeBr₃ will react to produce = (2.36 × 6) / 2 = 7.08 moles of NaBr

Therefore,

Thus, the theoretical yield of NaBr is 7.08 moles

Learn more about stoichiometry:

brainly.com/question/14735801

#SPJ1

7 0
2 years ago
Molar mass of a gas lab help?
LuckyWell [14K]

Answer:

The molar mass is determined by applying the Ideal Gas Law, PV = nRT, where P is the pressure (in atm), V is the volume (in L), n is the number of moles of gas, R is the universal gas constant (0.08206 L∙atm/mol∙K), and T is the temperature (in K).

Hope this helps! :)

3 0
3 years ago
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