Explanation:
Covalent compounds (solid, liquid, solution) do not conduct electricity. Metal elements and carbon (graphite) are conductors of electricity but non-metal elements are insulators of electricity. Ionic bonds are the electrostatic attraction between positive and negative ions.
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Answer:
The concentration of the NaOH solution CB = 0.251 M
Explanation:
The balanced equation of reaction is:
H2SO4 + 2NaOH ===> Na2SO4 + 2H2O
Using titration equation of formula
CAVA/CBVB = NA/NB
Where NA is the number of mole of acid = 1 (from the balanced equation of reaction)
NB is the number of mole of base = 2 (from the balanced equation of reaction)
CA is the concentration of acid = 0.1053 M
CB is the concentration of base = to be calculated
VA is the volume of acid = 17.88 mL
VB is the volume of base = 15.00mL
Substituting
0.1053×17.88/CB×15 = 1/2
Therefore CB =0.1053×17.88×2/15×1
CB= 0.251 M
Answer:
27.99 dm³
Explanation:
Applying
PV = nRT................ Equation 1
Where P = Pressure, V = Volume, n = number of mole, R = molar gas constant, T = Temperature.
From the question, we were aksed to find V.
Therefore we make V the subject of the equation
V = nRT/P................ Equation 2
Given: n = 1.31 moles, T = 37°C = 310K, P = 904 mmHg = (904×0.001316) = 1.1897 atm
Constant: R = 0.082 atm.dm³/K.mol
Substitute these values into equation 2
V = (1.31×310×0.082)/(1.1897)
V = 27.99 dm³
Answer:
Explanation:i joule is equal to 0.238902957619 calories so 1251 joules is equal to 298.87 calories divided by 25.0 degrees centigrade is equal to 11.95 calories divided by the 35.2 gram sample weight to get the calories per gram per degree centigrade would come to 0.3396 calories/gram degree centigrade. Presumably this, if correct, could be used to obtain the metal in question by consulting a chart or table with specific heats of various metals because they should always be the same specific heat for each metal.
Answer:
0.1035 M
Explanation:
Considering:
Sodium chloride will furnish Sodium ions as:
Given :
For Sodium chloride :
Molarity = 0.288 M
Volume = 3.58 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 3.58×10⁻³ L
Thus, moles of Sodium furnished by Sodium chloride is same the moles of Sodium chloride as shown below:
Moles of sodium ions by sodium chloride = 0.00103104 moles
Sodium sulfate will furnish Sodium ions as:
Given :
For Sodium sulfate :
Molarity = 0.001 M
Volume = 6.51 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 6.51 ×10⁻³ L
Thus, moles of Sodium furnished by Sodium sulfate is twice the moles of Sodium sulfate as shown below:
Moles of sodium ions by Sodium sulfate = 0.00001302 moles
Total moles = 0.00103104 moles + 0.00001302 moles = 0.00104406 moles
Total volume = 3.58 ×10⁻³ L + 6.51 ×10⁻³ L = 10.09 ×10⁻³ L
Concentration of sodium ions is:
<u>
The final concentration of sodium anion = 0.1035 M</u>