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Mariana [72]
2 years ago
7

How many grams are in 2 moles of Iodine?

Chemistry
2 answers:
Andre45 [30]2 years ago
7 0

Answer:

507.62 g

Explanation:

1 mole of iodine is = 253.81 g/mol^{-1\\}

253.81 x 2 = 507.62

Nady [450]2 years ago
6 0

Answer: 253.8

Explanation:

The molar mass of iodine is 126.904. Multiply that by two and you get approximately 253.8 grams in two moles.

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How many independent variables can you have in an experiment for it to be valid?
Lemur [1.5K]

Answer:

There are often not more than one or two independent variables tested in an experiment.

3 0
2 years ago
The ΔHsoln is _____.
nydimaria [60]

Sometimes negative sometimes positive, your answer is B!

4 0
3 years ago
Calculate the number of moles in 9.22 X 10^23 atom iron
sveta [45]
1.53 moles of Fe is your solution hope it helps!
7 0
2 years ago
A gas at constant temperature has a pressure of 404.6 kPa with a volume of 12 ml. If the volume changes to 43ml, what is the new
blagie [28]

Answer:

The answer is

<h2>112.912 kPa</h2>

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new pressure

P_2 =  \frac{P_1V_1}{V_2}  \\

404.6 kPa = 404600 Pa

From the question we have

P_2 =  \frac{404600 \times 12}{43}  =  \frac{4855200}{43}  \\  = 112911.6279... \\  = 112912

We have the final answer as

<h3>112.912 kPa</h3>

Hope this helps you

4 0
3 years ago
What is the ratio of lactic acid (Ka = 1.37x^10-4) to lactate in a solution with pH =4.29
hram777 [196]

Henderson–Hasselbalch equation is given as,

                                         pH  =  pKa  +  log [A⁻] / [HA]   -------- (1)

Solution:

Convert Ka into pKa,

                                         pKa  =  -log Ka

                                         pKa  =  -log 1.37 × 10⁻⁴

                                         pKa  =  3.863

Putting value of pKa and pH in eq.1,

                                         4.29  =  3.863 + log [lactate] / [lactic acid]

Or,

                   log [lactate] / [lactic acid]  =  4.29 - 3.863

                   log [lactate] / [lactic acid]  =  0.427

Taking Anti log,

                             [lactate] / [lactic acid]  =  2.673

Result:

           2.673 M  lactate salt when mixed with 1 M Lactic acid produces a buffer of pH = 4.29.

6 0
3 years ago
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