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3 years ago

Can someone plz help me with this one problem plzzzzz I’m being timed!!!

Chemistry

1 answer:

poizon [28]3 years ago

4 0

Answer:

answer : the penguins do

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10. A small gold nugget has volume of 0.87 cm3. What is its mass if the density of gold is 19.3 g/cm3?

bulgar [2K]

Answer:

16.791 grams

Explanation:

The density formula is:

$d=\frac{m}{v}$

Rearrange the formula for m, the mass. Multiply both sides of the equation by v.

$d*v=\frac{m}{v}*v$

$d*v=m$

The mass of the gold nugget can be found by multiplying the density and volume. The density is 19.3 grams per cubic centimeter and the volume is 0.87 cubic centimeters.

$d= 19.3 g/cm^3\\v-0.87 cm^3$

Substitute the values into the formula.

$m=d*v$

$m= 19.3 g/cm^3*0.87 cm^3$

Multiply. Note that the cubic centimeters, or cm³ will cancel each other out.

$m=19.3 g*0.87$

$m=16.791 g$

The mass of the gold nugget is 16.791 grams.

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3 years ago

This chart lists four elements from the periodic table.

aksik [14]

Iodine and Calcium is the correct answer

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3 years ago

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A nucleotide consists of a phosphate group, a pentose sugar, and a __________________, all linked together by covalent bonds. po

Genrish500 [490]

The correct answer would be the fourth option. A nucleotide consists of a phosphate group, a pentose sugar, and a nitrogen containing base that are all linked together by covalent bonds. Nucleotides are the monomer units of nucleic acids and is the basic unit of the DNA.

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3 years ago

Which electron in magnesium is most shielded from nuclear charge?

natali 33 [55]

An electron in the 3s orbital. The order of electron orbital energy levels starting from lowest to highest is as follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.

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3 years ago

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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