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3 years ago

Identify each element below, and give the symbols of the other elements in its group:

Chemistry

1 answer:

lozanna [386]3 years ago

7 0

Answer:

Answer in explanation

Explanation:

Argon has 18 electrons. So to get the element in question, we only need to add 18 to the number of the filled electrons.

a. Germanium, atomic number 32

Other group members:

Silicon Si , Carbon C , Tin Sn , Lead Pb and Flerovium Fl

b. Cobalt , atomic number 27

Other group members:

Rhodium Rh , Iridium Ir and Meitnerium Mt

c. Technetium , atomic number 43

Krypton is element 36

Other group members are :

Manganese Mn , Rhenium Re and Bohrium Bh

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= 25 X 5 = (use the correct number of sig figs)

Anton [14]

Answer:

1.25 *10^2

Explanation:

25*5 = 125

= 1.25 *10^2

5 0

3 years ago

The rate of a certain reaction is given by the following rate law: rate Use this information to answer the question below. At a

AleksAgata [21]

Answer:

Initial rate of the reaction when concentration of hydrogen gas is doubled will be $3.2\times 10^6 M/s$ .

Explanation:

$N_2+3H_2\rightarrow 2NH_3$

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

Initial rate of the reaction = R = $4.0\times 10^5 M/s$

$R = k\times [N_2][H_2]^3$

$4.0\times 10^5 M/s=k\times [N_2][H_2]^3$

The initial rate of the reaction when concentration of hydrogen gas is doubled : R'

$[H_2]'=2[H_2]$

$R'=k\times [N_2][H_2]'^3=k\times [N_2][2H_2]^3$

$R'=8\times k\times [N_2][H_2]^3$

$R'=8\times R=8\times 4\times 10^5 M/s=3.2\times 10^6 M/s$

Initial rate of the reaction when concentration of hydrogen gas is doubled will be $3.2\times 10^6 M/s$ .

7 0

3 years ago

determine mass of water formed when 12.5 L NH3(at298K and 1.50atm) is reacted with 18.9L of O2 (at 323K and 1.1atm)

sasho [114]

The mass of water formed is

calculation

Use the ideal gas equation to calculate the moles of NH3 and O2

that is Pv= n RT

where; P= pressure,

V= volume,

n = number of moles,

R=gas constant = 0.0821 l .atm/ mol.K

make n the formula of the subject by diving both side by RT

n = PV /RT

The moles of NH3

n= (1.50 atm x 12.5 L) /( 0.0821 L. atm /mol.k x 298 K) =0.766 moles

The moles of O2

=(1.1 atm x 18.9 L) / ( 0.0821 L. atm/ mol.k x 323 K) = 0.784 moles

write the reaction between NH3 and O2

4 NH3 + 5 O2 →4 No +6H2O

from equation above 0.766 moles of NH3 reacted to produce

0.766 x 6/4 =1.149 moles of H2O

0.784 moles of O2 reacted to produce 0.784 x 6/5=0.9408 moles of H20

since O2 is totally consumed, O2 is the limiting reagent and therefore the moles of H2O produced= 0.9408 moles

mass of H2O = moles x molar mass

from periodic table the molar mass of H2O = (1 x2)+16= 18 g/mol

mass = 18 g/mol x 0.9408 moles= 16.93 grams

3 0

3 years ago

A gaseous mixture contains 434.0 Torr H 2 ( g ) , 434.0 Torr H2(g), 389.9 Torr N 2 ( g ) , 389.9 Torr N2(g), and 77.9 Torr Ar (

weqwewe [10]

Answer:

H₂: 0.48, N₂: 0.43; Ar: 0.09

Explanation:

First of all, sum all the pressures to know the total pressure in the mixture.

434 Torr + 389.9 Torr + 77.9 Torr = 901.8 Torr

Mole fraction = Pressure gas / Total Pressure

Mole Fraction H₂: 434 Torr /901.8 Torr = 0.48

Mole Fraction N₂: 389.9 /901.8 Torr =0.43

Mole Fraction Ar: 77.9 /901.8 Torr = 0.09

Remember: SUM OF MOLE FRACTION = 1

8 0

3 years ago

A substance added in The final stages to remove sulphur from coal is

Deffense [45]

Answer:

calcium oxide is injected into the final stage of the scubber, wich then reacts with the sulfur dioxide to form calcium sulfite.

Explanation:

8 0

3 years ago