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klio [65]
3 years ago
10

Calculate the pH of a solution with an H+ ion concentration of 9.36 x 10-3 M ?

Chemistry
1 answer:
mel-nik [20]3 years ago
6 0

Answer:

The answer is 2.03

Explanation:

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

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

where H+ is the Hydrogen ion concentration

From the question we have

pH =  -  log(9.36 \times  {10}^{ - 3} )  \\  = 2.02872415...

We have the final answer as

<h3>2.03 </h3>

Hope this helps you

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Does 1 gram of phosphorus react with 6 grams of iodine to form 4 grams of phosphorus triodine in P4(s)+6I2(s)=4PI3(s)
mafiozo [28]

Answer:

No

Explanation:

One mole of P₄ react with six moles of I₂ and gives 4 moles of PI₃.

When one gram phosphorus and 6 gram of  iodine react they gives 8.234 g ram of PI₃ .

Given data:

Mass of phosphorus = 1 g

Mass of iodine = 6 g

Mass of  PI₃ = ?

Solution:

Chemical equation:

P₄ + 6I₂    →  4PI₃

Number of moles of P₄:

Number of moles = Mass /molar mass

Number of mole = 1 g / 123.9 g/mol

Number of moles  = 0.01 mol

Number of moles of I₂:

Number of moles  = Mass /molar mass

Number of moles = 6 g / 253.8 g/mol

Number of moles = 0.024 mol

Now we will compare the moles of PI₃ with I₂ and P₄.

                I₂              :              PI₃

                  6              :               4

                 0.024       :             4/6×0.024 = 0.02

                  P₄            :               PI₃

                 1                :                4

                 0.01          :               4 × 0.01 = 0.04  mol

The number of moles of PI₃ produced by I₂ are less it will be limiting reactant.

Mass of PI₃ = moles × molar mass

Mass of PI₃ = 0.02 mol × 411.7 g/mol

Mass of PI₃ =  8.234 g

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3 years ago
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Explanation:

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substitute: <span><span>t<span>1/2</span></span>=<span><span>ln(2)</span>k</span>→k=<span><span>ln(2)</span><span>t<span>1/2</span></span></span></span> Into the appropriate equation: <span>[A<span>]t</span>=[A<span>]0</span>∗<span>e<span>−kt</span></span></span> <span>[A<span>]t</span>=[A<span>]0</span>∗<span>e<span>−<span><span>ln(2)</span><span>t<span>1/2</span></span></span>t</span></span></span> <span>[A<span>]t</span>=(250.0 g)∗<span>e<span>−<span><span>ln(2)</span><span>3.823 days</span></span>(7.22 days)</span></span>=67.52 g</span>

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Explanation:

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