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stiks02 [169]
2 years ago
7

Calculate the mass of 29.8 mL of aluminum, which has a density of 2.00 g/mL.

Chemistry
1 answer:
MaRussiya [10]2 years ago
5 0

Answer:

The answer is

<h2>59.6 g </h2>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

Density of aluminum = 2.00 g/mL

volume = 29.8 mL

The mass is

mass = 2 × 29.8

We have the final answer as

<h3>59.6 g</h3>

Hope this helps you

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What is the percent of Cr in Cr2O3
IgorLugansk [536]
The percentage of Chromium in Chromium Oxide is calculated as follow,

Step 1:
           Calculate Molar mass of Cr₂O₃,
Cr = 51.99 u
O = 16 u
So,
     2(51.99) + 3(16) = 103.98 + 48 = 151.98 u

Step 2:
          Secondly divide molar mass of only chromium with total mass of Cr₂O₃ and multiply with 100.
i.e.
                   = \frac{103.98}{151.98} × 100
                   
                   = 68.41 %
So, the %age composition of chromium in chromium oxide is 68.41 %.
5 0
3 years ago
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which subatomic particle made out of protons neutrons and electrons can charge and still be considered the same element
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Answer:

Proton

Explanation:

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2 years ago
Codeine C18H21NO3 is a weak organic base. A 5.0 x 10^-3 M solution of codeine has a pH of 9.95. Calculate the value of Kb for th
arsen [322]

Answer:

Kb = 4.45\times10^{-7}\ mol/L

p^{Kb}=6.35

Explanation:

For a weak organic base, the formula to find p^{OH} is given by:

p^{OH}=p^{K_b}+\log c

where c is the concentration of base.

Here c= 5\times10^{-3}\ M

p^{H}=9.95\\p^{OH}=14-p^{H}=14-9.95=4.05

Substituting the above values in the formula,we get:

p^{k_b}=p^{OH}-\log c\\p^{k_b}=4.05-\log (5\times10^{-3})\\p^{K_b}=6.35\\K_b=$antilog 6.35=4.45\times10^{-7}\ mol/L

Hence:

Kb = 4.45\times10^{-7}\ mol/L

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7 0
3 years ago
Strontium chloride and sodium fluoride react to form strontium fluoride and sodium chloride, according to the reaction shown.
grin007 [14]

Answer:

4.07L of a 0.110M NaF are needed

Explanation:

Based on the reaction:

SrCl₂(aq)+2NaF(aq)⟶SrF₂(s)+2NaCl(aq)

<em>1 mole of strontium chloride react with 2 moles of NaF</em>

<em />

361mL of 0.620M SrCl₂ solution has:

0.361L ₓ (0.620mol / L) = 0.22382 moles SrCl₂.

Moles of NaF for a complete reaction must be:

0.22382 moles SrCl₂ ₓ (2 mol NaF / 1 mol SrCl₂) = <em>0.44764 moles of NaF</em>

If you have a solution of 0.110M NaF, the moles of NaF needed are:

0.44764 moles of NaF ₓ (1L / 0.110mol NaF) = <em>4.07L of a 0.110M NaF are needed</em>

<em></em>

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Elanso [62]

Answer:

the is is C

Explanation:

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3 years ago
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