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1 year ago

What is the number of electrons in an atom of scandium

1 answer:

4 0

The nuclear composition and electron configuration of an atom of Scandium 45 and the atomic number is 21 the most common isotype of this element is that nuclear consists of 21 protons and 24 neutrons Scandium is a transition metal group 3 period and the d-block of a periodic table

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What is the composition, in atom percent, of an alloy that contains a) 45.5 lbm of silver, b) 83.7 lbm of gold, and c) 6.3 lbm o

lyudmila [28]

Answer:

$\% atAg=44.6\%\\\% atAu=44.9\%\\\% atCu=10.5\%$

Explanation:

Hello,

In this case, for computing the atom percent, one must obtain the number of atoms of silver, gold and copper as shown below:

$atomsAg=45.5lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{107.87gAg}*\frac{6.022x10^{23}atomsAg}{1molAg}=1.15x10^{26}atomsAg\\atomsAu=83.7lbm*\frac{453.59g}{1lbm}*\frac{1molAu}{196.97gAu}*\frac{6.022x10^{23}atomsAu}{1molAu}=1.16x10^{26}atomsAu\\atomsCu=6.3lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{63.55gCu}*\frac{6.022x10^{23}atomsCu}{1molCu}=2.71x10^{25}atomsCu$

Thus, the atom percent turns out:

$\% atAg=\frac{1.15x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.6\%\\\% atAu=\frac{1.16x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.9\%\\\% atCu=\frac{2.71x10^{25}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =10.5\%$

Best regards.

4 0

2 years ago

Choose the formula for calculating percent composition.

kondor19780726 [428]

Answer:

The equation for percent composition is (mass of element/molecular mass) x 100.

please mark me as a brainlist

6 0

2 years ago

Determine how many carbon dioxide (CO2) molecules are produced if 8.45 x 1023 molecules of water (H2O) are produced 2 C2H6 (g) +

Andreas93 [3]

Ill get back to you on this

5 0

1 year ago

The rock in this picture shows foliation. Foliation can develop in a rock as a response to intense heat and pressure. This rock

MakcuM [25]

C) Metaphoric

This is the correct answer.

5 0

2 years ago

Read 2 more answers

A lithium flame has a characteristic red color due to emissions of wavelength 671 nm. What is the mass equivalence of 1 mol of p

e-lub [12.9K]

Answer:

1.98x10⁻¹² kg

Explanation:

The energy of a photon is given by:

E= hc/λ

h is Planck's constant, 6.626x10⁻³⁴ J·s

c is the speed of light, 3x10⁸ m/s

and λ is the wavelenght, 671 nm (or 6.71x10⁻⁷m)

E = 6.626x10⁻³⁴ J·s * 3x10⁸ m/s ÷ 6.71x10⁻⁷m = 2.96x10⁻¹⁹ J

Now we multiply that value by Avogadro's number, to calculate the energy of 1 mol of such protons:

1 mol = 6.023x10²³ photons

2.96x10⁻¹⁹ J * 6.023x10²³ = 1.78x10⁵ J

Finally we calculate the mass equivalence using the equation:

E=m*c²

m=E/c²

m = 1.78x10⁵ J / (3x10⁸ m/s)² = 1.98x10⁻¹² kg

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2 years ago