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uysha [10]
2 years ago
15

Which of the following options is correct?

Chemistry
1 answer:
Elena-2011 [213]2 years ago
8 0

Answer:

b. cellulose.

Explanation:

Cellulose is a polysaccharide consisting of a long chain of repeating glucose units (i.e., β-D-glucose units) that produce long unbranched chains. Cellulose can be considered as a natural polymer (i.e., a non-modified carbohydrate) because it is composed of repeated chains of β-D-glucose molecules stuck together. The cellulose polysaccharide is a principal component of plant cell walls.

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Can someone please answer this and if you dont know dont answer (:
anyanavicka [17]

Answer:  1- Alkali metals

                2- Noble gases

                3- Halogens

Explanation:

8 0
3 years ago
Read 2 more answers
What is the density of a 6cm3 cork that has a mass of 3g?
Andreyy89

Answer:

The density is 0,5 g/cm3

Explanation:

The density (δ) is the ratio between the mass and the volume of a compound:

δ=m/v= 3 g/6 cm3= <em>0,5 g/cm3</em>

5 0
3 years ago
Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to
Bess [88]

Answer:

Explanation:

To find the concentration; let's first compute the average density and the average atomic weight.

For the average density \rho_{avg}; we have:

\rho_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }

The average atomic weight is:

A_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }

So; in terms of vanadium, the Concentration of iron is:

C_{Fe} = 100 - C_v

From a unit cell volume V_c

V_c = \dfrac{n A_{avc}}{\rho_{avc} N_A}

where;

N_A = number of Avogadro constant.

SO; replacing V_c with a^3 ; \rho_{avg} with \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} } ; A_{avg} with \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} } and

C_{Fe} with 100-C_v

Then:

a^3 = \dfrac   { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) }    {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big)  }

a^3 = \dfrac   { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) }    {N_A  \Big (\dfrac{100 \times \rho_{Fe} \times  \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big)  }

a^3 = \dfrac   { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) }    {N_A  \Big (\dfrac{100 \times \rho_{Fe} \times  \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big)  }

Replacing the values; we have:

(0.289 \times 10^{-7} \ cm)^3 = \dfrac{2 \ atoms/unit \ cell}{6.023 \times 10^{23}} \dfrac{ \dfrac{100 (50.94 \g/mol) (55.84(g/mol)} { 100(50.94 \ g/mol) - C_v(50.94 \ g/mol) + C_v (55.84 \ g/mol)   }   }{ \dfrac{100 (7.84 \ g/cm^3) (6.0 \ g/cm^3 } { 100(6.0 \ g/cm^3) - C_v(6.0 \ g/cm^3) + C_v (7.84 \ g/cm^3)   } }

2.41 \times 10^{-23} = \dfrac{2}{6.023 \times 10^{23} }  \dfrac{ \dfrac{100 *50*55.84}{100*50.94 -50.94 C_v +55.84 C_v} }{\dfrac{100 * 7.84 *6}{600-6C_v +7.84 C_v} }

2.41 \times 10^{-23} (\dfrac{4704}{600+1.84 C_v})=3.2 \times 10^{-24} ( \dfrac{284448.96}{5094 +4.9 C_v})

\mathbf{C_v = 9.1 \ wt\%}

4 0
3 years ago
PLEASE PLEASE HURRRY???!!!!!!!!!
KiRa [710]

answer a) the existence of isomers is a reason for the large number of carbon compounds

4 0
2 years ago
35 cm3 of a metal has a mass of 121 grams. whats is its density?
dusya [7]

Answer:

The density of the metal is 3.457 g/cm³.

Explanation:

Given,

mass of metal = 121 grams

Volume = 35 cm³.

Density = ?

Density of the metal can be found by using the formula

Density = Mass/Volume

Substituting the values,

Density = 121/35 = 3.457 g/cm³.

5 0
2 years ago
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