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olya-2409 [2.1K]
3 years ago
12

Please help me with this question. image attached

Chemistry
1 answer:
telo118 [61]3 years ago
7 0

Answer:

A, D,E

Explanation:

I think thats right. Cuz losing electrons makes you cation, gaining electron by a neutral atom makes you negative, and giving up electrons is synonymous to acting like a "metal"

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Suppose that 25,0 mL of a gas at 725 mmHg and 298K is converted to
posledela

The new volume : 21.85 ml

<h3>Further explanation</h3>

Given

V1=25,0 ml

P1=725 mmHg

T1=298K is converted to

T2=273'K

P2=760 mmHg atm

Required

V2

Solution

Combined gas law :

\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}

Input the value :

V2=(P1.V1.T2)/(P2.T1)

V2=(725 x 25 ml x 273)/(760 x 298)

V2=21.85 ml

5 0
3 years ago
What happens to dissolving rate when you increase KE?
TEA [102]
In general, solubility increases with temperature. When you increase the temperature of a solvent, you increase the kinetic energy (or energy of movement) of the molecules, and this greater energy helps dissolve more of the solute molecules.
4 0
3 years ago
Calculate the mole fraction of cai2 in an aqueous solution prepared by dissolving 0.400 moles of cai2 in 850.0 g of water.
alukav5142 [94]
1) Formulas:

a) mole fraction of component 1, X1

X1 =  number of moles of compoent 1 / total number of moles

b) Molar mass = number grams / number of moles => number of moles =  number of grams / molar mass


2) Application

Number of moles of CaI2 = 0.400

Molar mass of water = 18.0 g/mol

Number of moles of water: 850.0 g / 18.0 g/mol = 47.22 mol

Total number of moles = 0.400 + 47.22 =47.62

Molar fraction of CaI2 = 0.400 / 47.62 = 0.00840
7 0
3 years ago
What is the enthalpy of combustion (per mole) of C4H10 (g)? 
Artemon [7]
The balanced chemical reaction for the complete combustion of C4H10 is shown below:

                    C4H10 + (3/2)O2 --> 4CO2 + 5H2O

The enthalpy of formation are listed below:
          C4H10: -2876.9 kJ/mol
              O2:   none (because it is pure substance)
             CO2: -393.5 kJ/mol
             H2O: -285.8 kJ/mol

The enthalpy of combustion is computed by subtracting the total enthalpy formation of the reactants from that of the products.

               ΔHc = (4)(-393.5 kJ/mol) + (5)(-285.8 kJ/mol) - (-2876.9 kJ/mol)
                       = -<em>126.1 kJ</em>

Thus, the enthalpy of combustion of the carbon is -126.1 kJ. 
5 0
3 years ago
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vovangra [49]
Do not trust people that share links these people hack and put viruses. to your phone/computer’s


DO NOT TRUST THESE LINKS
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3 years ago
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