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Airida [17]
3 years ago
6

How many grams of H2 will be produced, given 27.4 g of Al to start with?

Chemistry
1 answer:
pishuonlain [190]3 years ago
6 0
3.07g H2

27.4/26.98/2x3x1.01x2=3.07
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Please if any one know any question help me please
pashok25 [27]
Hhh hhbththrbvevevevevevr
7 0
2 years ago
Calculate the number of moles in 25.0 g of each of the
mamaluj [8]

Answer:

A. 6.25moles

B. 0.78moles

C. 0.32moles

D. 0.15moles

E. 0.43moles

Explanation:

use

  • n = mass/molar mass

A. He

  • 25/4 = 6.25moles

B. O2

  • 25/32 = 0.78moles

C. Al(OH)3

  • 27+ 3(17) = 78
  • 25/78 = 0.32 moles

D. GaS3

  • 70+3(32) = 166
  • 25/166 = 0.15moles

E. C4H10

  • 4(12)+10(1) =58
  • 25/58 = 0.43moles
5 0
2 years ago
Definition of periodic trends
Dmitriy789 [7]

Answer:

.Periodic trends are specific patterns in the properties of chemical elements that are revealed in the periodic table of elements. Major periodic trends include electronegativity, ionization energy, electron affinity, atomic radii, ionic radius, metallic character, and chemical reactivity.

7 0
3 years ago
Convert 32.56 km/hr into ft/hr
Gekata [30.6K]
Note that
1 m = 3.2808 ft

Therefore
1 km = 3280.8 ft
and
32.56 \,  \frac{km}{h} = (32.56 \,  \frac{km}{h})*(3280.8 \,  \frac{ft}{km}) =  1.0682 \, \times 10^{5} \, \frac{ft}{h}

Answer: 1.0682 x 10⁵ ft/hr

8 0
3 years ago
A chemist wishing to do an experiment requiring 47-Ca2+ (half-life = 4.5 days) needs 5.0 μg of the nuclide. What mass of 47-CaCO
NARA [144]

Answer:

5.8μg

Explanation:

According to the rate or decay law:

N/N₀ = exp(-λt)------------------------------- (1)

Where N = Current quantity,  μg

            N₀ = Original quantity, μg

             λ= Decay constant day⁻¹

              t =  time in days

Since the half life is 4.5 days, we can calculate the  λ from (1) by  substituting N/N₀ = 0.5

0.5 = exp (-4.5λ)

ln 0.5  = -4.5λ

-0.6931 = -4.5λ

λ =   -0.6931 /-4.5

  =0.1540 day⁻¹

Substituting into (1)  we have :

N/N₀ = exp(-0.154t)----------------------------- (2)

To receive 5.0 μg of the nuclide with a delivery time of 24 hours or 1 day:

N = 5.0 μg

N₀ = Unknown

t = 1 day

Substituting into (2) we have

[5/N₀]   = exp (-0.154 x 1)

    5/N₀        = 0.8572

N₀  =  5/0.8572

     =    5.8329μg

    ≈     5.8μg

The Chemist must order 5.8μg  of 47-CaCO3

6 0
3 years ago
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