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mel-nik [20]
3 years ago
15

How many moles are equal to 1.3 x 10^24 atoms of aluminum?

Chemistry
1 answer:
emmasim [6.3K]3 years ago
7 0

Answer:

Explanation: well I think 6.02 x 10^23 is 1 mole. So do 1/6.02 x 10^23 = x/1.3 x 10^24

You might be interested in
Which chemical equation represents a synthesis reaction?A.HCl + LiOH → LiCl + H2OB.2Na + MgCl2 → 2NaCl + Mg.H2CO3 → H2O + CO2D.C
Zina [86]

Answer:

D

Explanation:

Just did it on A P E X

6 0
3 years ago
Calculate the rate of dissolution (dM/dt) of relatively hydrophobic drug particles with a surface area of 2.5×103 cm2 and satura
Crank

Answer:

\large \boxed{\text{1.22 mg/s}}

Explanation:

We can use the Noyes-Whitney equation to calculate the rate of dissolution.

\dfrac{\text{d}M}{\text{d}t} = \dfrac{DA(C_{s} - C)}{d}

Data:

D = 1.75 × 10⁻⁷ cm²s⁻¹

A = 2.5 × 10³ cm²

Cₛ = 0.35 mg/mL

C = 2.1 × 10⁻⁴ mg/mL

d = 1.25 µm

Calculations:

Cₛ - C = (0.35 - 2.1 × 10⁻⁴) mg·cm⁻³ = 0.350 mg·cm⁻³

d = 1.25 µm = 1.25 × 10⁻⁶ m = 1.25 × 10⁻⁴ cm

\dfrac{\text{d}M}{\text{d}t} = \dfrac{(1.75 \times 10^{-7} \text{cm}^{2}\text{s}^{-1})(2.5 \times 10^{3} \text{ cm}^{2})(0.350\text{ mg$\cdot$cm$^{-3}$})}{1.25 \times 10^{-4} \text{ cm}} = \textbf{1.22 mg/s}\\\\\text{The rate of dissolution is $\large \boxed{\textbf{1.22 mg/s}}$}

8 0
3 years ago
Two perfumes are released at the same time. If one is standing 7.5 m from the point of release. Perfume A (molar mass 275 g/mol)
natta225 [31]
According to Graham's Law of Diffusion," Diffusion of Gas is inversely proportional to square root of its Molecular Mass or Density".

                          rᵇ/rᵃ  =  \sqrt{da/db}
Or,
                          rᵇ/rᵃ  =  \sqrt{Ma/Mb}     ----- (1)
As, 
                          Ma  =  275 g/mol

                          Mb  =  205 g/mol

Putting Values in eq.1,

                          rᵇ/rᵃ  =  \sqrt{275/205}

                          rᵇ/rᵃ  =  1.15

Result:
          Perfume B will diffuse 1.15 times faster than Perfume A. Hence, Perfume B will be first smelled by the person.
6 0
3 years ago
Rank the following fertilizers in decreasing order of mass percentage of nitrogen:
charle [14.2K]
<h3>Answer:</h3>

        NH₃ > NH₄NO₃ > (NH₄)₂HPO₄ > (NH₄)₂SO₄ > KNO₃ > (NH₄)H₂PO₄

<h3>Soution:</h3>

In (NH₄)₂HPO₄:

Mass of Nitrogen  =  N × 2  =  14 × 2  =  28 g.mol⁻¹

Molar Mass of (NH₄)₂HPO₄  =  132.06 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of (NH₄)₂HPO₄ × 100

Mass %age  =  28 g.mol⁻¹ / 132.06 g.mol⁻¹ × 100

Mass %age  =  21.20 %

In (NH₄)₂SO₄:

Mass of Nitrogen  =  N × 2  =  14 × 2  =  28 g.mol⁻¹

Molar Mass of (NH₄)₂SO₄  =  132.14 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of (NH₄)₂SO₄ × 100

Mass %age  =  28 g.mol⁻¹ / 132.14 g.mol⁻¹ × 100

Mass %age  =  21.18 %

In KNO₃:

Mass of Nitrogen  =  N × 1  =  14 × 1  =  14 g.mol⁻¹

Molar Mass of KNO₃  =  101.10 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of KNO₃ × 100

Mass %age  =  14 g.mol⁻¹ / 101.10 g.mol⁻¹ × 100

Mass %age  =  13.84 %

In (NH₄)H₂PO₄:

Mass of Nitrogen  =  N × 1  =  14 × 1  =  14 g.mol⁻¹

Molar Mass of (NH₄)H₂PO₄  =  115.03 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of (NH₄)H₂PO₄ × 100

Mass %age  =  14 g.mol⁻¹ / 115.03 g.mol⁻¹ × 100

Mass %age  =  12.17 %

In NH₃:

Mass of Nitrogen  =  N × 1  =  14 × 1  =  14 g.mol⁻¹

Molar Mass of NH₃  =  132.14 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of NH₃ × 100

Mass %age  =  14 g.mol⁻¹ / 17.03 g.mol⁻¹ × 100

Mass %age  =  82.20 %

In NH₄NO₃:

Mass of Nitrogen  =  N × 2  =  14 × 2  =  28 g.mol⁻¹

Molar Mass of NH₄NO₃  =  80.04 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of NH₄NO₃ × 100

Mass %age  =  28 g.mol⁻¹ / 80.04 g.mol⁻¹ × 100

Mass %age  =  34.98 %

5 0
3 years ago
How much heat does it take to turn water from -30 degreesC to 120 degreesC?
natka813 [3]

Answer:

110 degrees C

Explanation:

4 0
2 years ago
Read 2 more answers
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