All categories

2 years ago

Which statement best describes the atomic structure of metals

Chemistry

1 answer:

Vikki [24]2 years ago

6 0

Answer:

Closely packed cations with loosely held valence electrons

Explanation:

Pure metals contain so-called metallic bonds. These metallic bonds are often referred to as sea of electrons. Metals consist of closely packed cations (and, hence, that's why they have a high density) with loosely held valence electrons (this is the reason why metals tend to lose their electrons rather than gain, besides, it's also the reason why the first ionization energy for the metals is relatively low).

You might be interested in

HELP! BEST ANSWER GET A FOLLOW AND BRAINLIEST

DiKsa [7]

The answer I got was : 52.2

4 0

2 years ago

Read 2 more answers

What is the mass of 3.20x10^23 formula units of iron (III) oxide (Fe2O3)?

yaroslaw [1]

The mass of iron (III) oxide (Fe2O3) : 85.12 g

<h3>Further explanation</h3>

Given

3.20x10²³ formula units

Required

The mass

Solution

1 mole = 6.02.10²³ particles

Can be formulated :

N = n x No

N = number of particles

n = mol

No = 6.02.10²³ = Avogadro's number

mol of Fe₂O₃ :

$\tt n=\dfrac{3.2.10^{23}}{6.02.10^{23}}=0.532$

mass of Fe₂O₃ (MW=160 g/mol)

$\tt mass=mol\times MW=0.532\times 160=85.12~g$

4 0

2 years ago

Please help, with step by step work

natali 33 [55]

$\qquad$ ☀️ $\pink{\bf{ {Answer = \: \: 85.57g }}}$

Molar mass of $\bf Cu_2O$

$\qquad$ $\twoheadrightarrow\sf 63.546 \times 2 +16$

$\qquad$ $\pink{\twoheadrightarrow\bf 143.092 g}$

<u>As we know</u>–

1 mol = $\bf 6.02×10^{23}$ formula units

1 mol $\bf Cu_2O$ = 143.092 g = $\bf 6.02×10^{23}$ formula units

Henceforth –

$\bf 3.60×10^{23}$ formula units $\bf Cu_2O$ –

$\qquad$ $\sf :\implies \dfrac{143.092 \times3.60×10^{23 }}{6.02×10^{23}}$

$\qquad$ $\sf :\implies \dfrac{143.092 \times3.60×\cancel{10^{23 }}}{6.02×\cancel{10^{23}}}$

$\qquad$ $\pink{:\implies\bf 85.57 g}$

5 0

2 years ago

A 72.0 mL aliquot of a 1.40 M solution is diluted to a total volume of 248 mL. A 124 mL portion of that solution is diluted by a

Aliun [14]

Answer: 0.20 M

Explanation:

According to the dilution law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = 1.40 M

$V_1$ = volume of stock solution = 72.0 ml

$M_2$ = molarity of diluted solution = m

$V_2$ = volume of diluted solution = 248 ml

$1.40\times 72.0=m\times 248$

$m=0.41M$

Now 124 mL portion of this prepared solution is diluted by adding 133 mL of water.

According to the dilution law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = 0.41 M

$V_1$ = volume of stock solution = 124 ml

$M_2$ = molarity of diluted solution = m

$V_2$ = volume of diluted solution = (124 +133) ml = 257 ml

$0.41\times 124=m\times 257$

$m=0.20M$

Thus the final concentration of the solution is 0.20 M.

8 0

3 years ago

If you want to determine the actual salicylate concentration for a sample that gives a greater absorption than the highest conce

Anarel [89]

Answer: The concentration of unknown sample is 1.8 mg/ml

Explanation:

According to the dilution law,

$M_1V_1=M_2V_2$

where,

$M_1$ = concentration of stock solution = ?

$V_1$ = volume of stock solution = 0.25 ml

$M_1$ = concentration of diluted solution = 0.45 mg/ml

$V_1$ = volume of diluted solution = (0.25+0.75) ml = 1.00 ml

Putting in the values we get:

$M_1\times 0.25=0.45\times 1.00$

$M_1=1.8mg/ml$

Therefore, concentration of unknown sample is 1.8 mg/ml

3 0

2 years ago