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Elena L [17]
3 years ago
11

Is palladium a metal nonmetal or metalloid

Chemistry
2 answers:
Dima020 [189]3 years ago
8 0

Answer:

It is a rare and lustrous silvery-white metal

Explanation:

kap26 [50]3 years ago
7 0

Answer: metal i believe

Explanation:

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A. How many moles of copper equal 8.00 × 109 copper atoms?
BARSIC [14]

Answer:

Explanation:

a ) one mole = 6.02 x 10²³ atoms

no of moles in given no of atoms

= 8 x 10⁹ / 6.02 x 10²³

= 1.329 x 10⁻¹⁴ moles .

b ) one mole of calcium = 40 gram

102 .5 g calcium

= 102 .5 / 40 moles

= 2.5625 moles

c )

no of moles in 5.04 g lead = 5.04 / 207

= 2.4347 x 10⁻² moles

= 2.4347 x 10⁻²x 6.02 x 10²³ no of atoms of lead

= 14.6568 x 10²¹ no of atoms .

d)  

one mole = 6.02 x 10²³ atoms

2.85 mole = 17.157 x 10²³ atoms .

e )

moles of fluorine gas = 1.08 x 10³ / 6.02 x 10²³

= .1794 x 10⁻²⁰ moles

mass in grams =  .1794 x 10⁻²⁰ x 38

= 6.8172 x 10⁻²⁰ grams

f )

no of moles in .584 g of benzene = .584 / 78

= 7.487 x 10⁻³ moles

no of molecules = 6.02 x 10²³ x  7.487 x 10⁻³

= 45.07 x 10²⁰ molecules .

g )

moles of atoms = 5.09 x 10⁹ / 6.02 x 10²³

= .8455 x 10⁻¹⁴ moles

mass in gram = .8455 x 10⁻¹⁴ x 1

=  .8455 x 10⁻¹⁴ g

h )

.45 moles of Ca₃PO₄ = .45 x 6.02 x 10²³ molecules

= 2.709 x 10²³ molecules of Ca₃PO₄

no of atoms of Ca = 3 x 2.709 x 10²³

= 8.127 x 10²³ atoms of Ca .

7 0
4 years ago
The symbol for gold is<br> a. go.<br> b. g.<br> c. au.<br> d. au.
Zolol [24]
Au is the symbol for gold. It comes from Latin word for gold, which is Aurunum. Its atomic number is 79. 
7 0
3 years ago
Read 2 more answers
Н
AysviL [449]

Answer:

A 3,3-dimethylhexane

8 0
4 years ago
Calculate the energy, in joules, required to ionize a hydrogen atom when its electron is initially in the n =2 energy level. The
qaws [65]

Answer:

E_{ionization}=5.45\times 10^{-19}\ J

Explanation:

E_n=-2.18\times 10^{-18}\times \frac{1}{n^2}\ Joules

For transitions:

Energy\ Difference,\ \Delta E= E_f-E_i =-2.18\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

\Delta E=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

So, n_i=2 and n_f=\infty (As the hydrogen has to ionize)

Thus,

\Delta E=2.18\times 10^{-18}(\frac{1}{2^2} - \dfrac{1}{{\infty}^2})\ J

\Delta E=2.18\times 10^{-18}(\frac{1}{2^2})\ J

E_{ionization}=5.45\times 10^{-19}\ J

4 0
3 years ago
What STARTING element undergoes alpha decay to yield uranium-234 (U) as a PRODUCT?
nadezda [96]
Uranium-238 is the element that undergoes decay to yield uranium-234
3 0
3 years ago
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