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wlad13 [49]
3 years ago
15

1s22s22p63523p645230104p5 Which element is this? answer : bromine

Chemistry
1 answer:
Svetradugi [14.3K]3 years ago
8 0

Answer:

indeed it 1s22s22p63523p645230104p5 is bromine

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Which statement best describes covalent bonding?
almond37 [142]

Answer:

Option C. Electrons are shared between two atoms

Explanation:

Covalent bonding is a type of bonding which exist between two non metals.

In this bonding, electrons are shared between the two atoms involved in order to attain a stable octet configuration.

This can be seen when hydrogen atom combine with chlorine atom to form hydrogen chloride as shown below:

H + Cl —> HCl

Hydrogen has 1 electron in it's outmost shell and it requires 1 electron to attain a stable configuration.

Chlorine has 7 electrons in it's outmost shell and requires 1 electron to attain a stable configuration.

During bonding, both hydrogen and chlorine will contribute 1 electron each to form bond, thereby attaining a stable configuration. The bond formed in this case is called covalent bond as both atoms involved shared electron to attain a stable configuration.

3 0
3 years ago
List at least 3 factors that can influence a body's temperature after death.
Slav-nsk [51]

multiply answers

ventilation

humidity

insulation

surface temperature

Explanation:

ventilation-well ventilated room could increase the body's cooling process

humidity-being in a humid location could cool the body at a slower rate

insulation-a body wrapped in something will cool slower then one not

surface temperature-a body on a hit surface will cool slower than one on a cold surface

3 0
3 years ago
A certain volume of a 0.80 m solution contains 4.3 g of a salt. what mass of the salt is present in the same volume of a 2.2 m s
Nat2105 [25]
In a 2.2 m solution with the same volume there would be 11.825 g of a salt
6 0
4 years ago
Use Q = mcAT
mr Goodwill [35]

Answer:

1).....for the specific heat capacity(c) of water is 4200kg/J°C..

....guven mass(m)=320g(0.32kg)

...change in temperature(ΔT) =35°C

from the formula

Q=mcΔT

Q=0.32Kg x 4200kg/J°C x 35°C

Q=47,040Joules

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
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