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MaRussiya [10]
3 years ago
7

How many grams of tris (mw 121.1) would you need to prepare 100 ml of a 100 mm tris solution? _________ grams?

Chemistry
1 answer:
Tems11 [23]3 years ago
7 0

Molarity = (Mass/ molar mass) x (1/ volume of solution in Litres)

Mass = Molarity x molar mass x  volume of solution in Litres

Molarity of Tris = 100 mM = 0.1 M

volume of Tris sol. = 100 mL = 0.1 L

molar mass of Tris = 121.1 g/mol

Hence,

mass of Tris = Molarity of Tris x molar mass ofTris x volume of Tris solution

= 0.1 M x 121.1 g/mol x 0.1 L

= 1.211 g

mass of Tris = 1.211 g


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What is the OH- of {H+} = 4.0 x 10 to the power of -8
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At standard room temperature, [{\rm OH^{-}] \approx 2.5 \times 10^{-7}\; \rm M when [{\rm H^{+}] = 4.0 \times 10^{-8}\; \rm M.

Explanation:

The following equilibrium goes on in water:

{\rm H_{2}O}\, (l) \rightleftharpoons {\rm H^{+}}\, (aq) + {\rm OH^{-}}\, (aq).

The forward reaction is known as the self-ionization of water. The ionization constant of water, K_{\rm w}, gives the equilibrium position of this reaction:

K_{\rm w} = [{\rm H^{+}] \cdot [{\rm OH^{-}}].

At standard room temperature (25\; {\rm ^{\circ}C}), K_{\rm w} \approx 10^{-14}. Also, [{\rm H^{+}}] = 4.0 \times 10^{-8}\; \rm mol \cdot L^{-1}. Substitute both values into the equation and solve for [{\rm OH^{-}}].

\begin{aligned} {[}{\rm OH^{-}}{]} &= \frac{K_{\rm w}}{[{\rm H^{+}}]} \\ &\approx \frac{10^{-14}}{4.0 \times 10^{-8}} = 2.5 \times 10^{-7}\end{aligned}.

In other words, in an aqueous solution at standard room temperature, [{\rm OH^{-}] \approx 2.5 \times 10^{-7}\; \rm M when [{\rm H^{+}] = 4.0 \times 10^{-8}\; \rm M.

5 0
3 years ago
If a gas occupies 4600 mL at 0.9 atm and 195°C, what is the new volume in ml
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Answer:

The new volume is 2415 mL

Explanation:

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases.

Boyle's law says that the volume occupied by a given gas mass at constant temperature is inversely proportional to the pressure and is expressed mathematically as:

P * V = k

Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

\frac{V}{T} =k

Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature is decreased, the gas pressure decreases. This can be expressed mathematically in the following way:

\frac{P}{T} =k

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

\frac{P*V}{T} =k

Having two different states, an initial state and an final state, it is true:

\frac{P1*V1}{T1} =\frac{P2*V2}{T2}

In this case:

  • P1= 0.9 atm
  • V1=4,600 mL= 4.6 L (being 1 L=1,000 mL)
  • T1= 195 °C= 468 °K (being 0°C=273°K)

The final state 2 is in STP conditions:

  • P2= 1 atm
  • V2= ?
  • T2= 0°C= 273 °K

Replacing:

\frac{0.9 atm*4.6L}{468K} =\frac{1 atm*V2}{273K}

Solving:

V2=\frac{0.9 atm*4.6L}{468K}*\frac{273K}{1 atm}

V2= 2.415 L =2,415 mL

<u><em>The new volume is 2415 mL</em></u>

6 0
3 years ago
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