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Arte-miy333 [17]
2 years ago
13

What is the pH of a solution with a hydrogen ion concentration of 4.5 x 10-6 M ?

Chemistry
1 answer:
Fittoniya [83]2 years ago
6 0

Answer:

<h2>5.35 </h2>

Explanation:

The pH of a solution can be found by using the formula

pH = - log [ {H}^{+} ]

We have

pH =  -  log(4.5 \times  {10}^{ - 6} )  \\  = 5.346787...

We have the final answer as

<h3>5.35 </h3>

Hope this helps you

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PLSSSSS HELP I DONT GET THIS PROBLEMMMM
Aleks [24]

Answer:

C. 7370 joules.

Explanation:

There is a mistake in the statement. Correct form is described below:

<em>Using the above data table and graph, calculate the total energy in Joules required to raise the temperature of 15 grams of ice at -5.00 °C to water at 35 °C. </em>

The total energy needed to raise the temperature is the combination of latent and sensible heats, all measured in joules, and represented by the following model:

Q = m\cdot [c_{i} \cdot (T_{2}-T_{1})+L_{f} + c_{w}\cdot (T_{3}-T_{2})] (1)

Where:

m - Mass of the sample, in grams.

c_{i} - Specific heat of ice, in joules per gram-degree Celsius.

c_{w} - Specific heat of water, in joules per gram-degree Celsius.

L_{f} - Latent heat of fusion, in joules per gram.

T_{1} - Initial temperature of the sample, in degrees Celsius.

T_{2} - Melting point of water, in degrees Celsius.

T_{3} - Final temperature of water, in degrees Celsius.

Q - Total energy, in joules.

If we know that m = 15\,g, c_{i} = 2.06\,\frac{J}{g\cdot ^{\circ}C}, c_{w} = 4.184\,\frac{J}{g\cdot ^{\circ}C}, L_{f} = 334.72\,\frac{J}{g}, T_{1} = -5\,^{\circ}C, T_{2} = 0\,^{\circ}C and T_{3} = 35\,^{\circ}C, then the final energy to raise the temperature of the sample is:

Q = (15\,g)\cdot \left[\left(2.06\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (5\,^{\circ}C)+ 334.72\,\frac{J}{g} + \left(4.184\,\frac{J}{g\cdot ^{\circ}C}\right)\cdot (35\,^{\circ}C) \right]

Q = 7371.9\,J

Hence, the correct answer is C.

8 0
3 years ago
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3CaCl2 +2Na3PO4-->6NaCl +Ca3(PO4)2
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from  the  equation  above  the  ratio  of  CaCl2  to  Ca3(PO4)2  is  3:1  therefore  the  moles  of Ca3(PO4)2  is 0.809/3=0.268moles
mass  is  therefore  0.268  x310.18(R.F.M of   Ca3(PO4)2 ) =83.23grams
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Drag always opposes forward motion or slows you down.

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endothermic reaction

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