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Komok [63]
3 years ago
6

When 4.5 mol al react with 11.5 mol hcl, what is the limiting reactant, and how many moles of h2 can be formed?

Chemistry
1 answer:
ahrayia [7]3 years ago
5 0
1°/ . 2 Al + 6 HCl → 2 AlCl3 + 3 H2 

<span>k1 = n(Al) / 2 = 4,5 / 2 = 2,25 </span>
<span>k2 = n(HCl) / 6 = 11,5 / 6= 1,92 </span>

<span>k2 < k1 ==> HCl is the limiting reactant </span>

<span>6 mol of HCl ---> 2 mol of H2 </span>
<span>11,5 mol of HCl ---> 3,83 mol of H2 </span>
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Rank the elements according to highest ionization energy:<br><br> Be, C, O, Ne, B, Li, F, N
aleksklad [387]
Given:
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C   - Carbon      - 11,2603
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B   - Boron          -   8,298
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Hurry please!
Pani-rosa [81]

Answer : The mass of of water present in the jar is, 298.79 g

Solution : Given,

Mass of barium nitrate = 27 g

The solubility of barium nitrate at 20^oC is 9.02 gram per 100 ml of water.

As, 9.02 gram of barium nitrate present in 100 ml of water

So, 27 gram of barium nitrate present in \frac{27g}{9.02g}\times 100ml=299.33ml of water

The volume of water is 299.33 ml.

As we know that the density of water at 20^oC is 0.9982 g/ml

Now we have to calculate the mass of water.

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}

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Therefore, the mass of of water present in the jar is, 298.79 g

5 0
3 years ago
Astroturf is a durable artificial surface used to cover athletic fields. A soccer field 0.06214- mile-long by 253 ft wide is cov
love history [14]

Answer:

The weight of the Astroturf is 179,684.31 Newtons.

Explanation:

Length of a soccer field = 0.06214 mile = 328.0992 feet

(1 mile = 5280 feet)

Breadth of a soccer field  = 253 feet

Length of a Astroturf which soccer field is to be covered, l = 328.0992 feet

Breadth of a Astroturf which soccer field is to be covered ,b = 253 feet

Thickness of a Astroturf with which soccer field is to be covered = h

h = ½ inch = 0.5 inch = 0.041665 feet

(1 inches = 0.08333 feet)

Volume of the Astroturf ,V= l × b × h

V=328.0992 ft\times 253 ft\times 0.041665 ft=3,458.574 ft^3

Mass of the Astroturf = m

Density of the Astroturf = d = 187 oz/ft^3

d=\frac{m}{V}

m=d\times V= 187 oz/ft^3\times 3,458.574 ft^3=646,753.35 oz

1 oz = 0.0283495 kg

m=646,743.35 oz=646,743.35\times 0.0283495 kg=18,335.13 kg

Weight of the Astroturf = W

W = mg

=W=18,335.13 kg\times 9.8 m/s^2=179,684.31 N

The weight of the Astroturf is 179,684.31 Newtons.

8 0
3 years ago
Determine how many mm of Hg are equal to 5.3 atm of pressure
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4028 mm of Hg...........
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