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lys-0071 [83]
3 years ago
9

Suggest why it might be difficult and dangerous to test sodium for electrical conductivity?​

Chemistry
1 answer:
lukranit [14]3 years ago
8 0

Answer:

It might cause fire or extreme smoke.

Explanation:

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Calculate the millimoles of solute in 1.88 L of a 0.00713 M NaCN solution. millimoles:
konstantin123 [22]

Answer:

13.4 milimoles.

Explanation:

The following data were obtained from the question:

Volume = 1.88 L

Molarity = 0.00713 M

Millimoles of NaCN =?

Next, we shall determine the number of mole NaCN in the solution. This can be obtained as follow:

Molarity = mole /Volume

Volume = 1.88 L

Molarity = 0.00713 M

Mole of NaCN =?

Molarity = mole /Volume

0.00713 = moles of NaCN /1.88

Cross multiply

Mole of NaCN = 0.00713 × 1.88

Mole of NaCN = 0.0134 mole

Finally, we shall convert 0.0134 mole to Millimoles. This can be obtained as follow:

1 mole = 1000 millimoles

Therefore,

0.0134 mole = 0.0134 × 1000

0.0134 mole = 13.4 milimoles

Therefore, the millimoles of the solute, NaCN in the solution is 13.4 milimoles

7 0
4 years ago
Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or
Ahat [919]

Answer :

Oxidation number or oxidation state : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

Rules for Oxidation Numbers are :

  • The oxidation number of a free element is always zero.
  • The oxidation number of a monatomic ion equals the charge of the ion.
  • The oxidation number of  Hydrogen (H)  is +1, but it is -1 in when combined with less electronegative elements.
  • The oxidation number of  oxygen (O)  in compounds is usually -2.
  • The oxidation number of a Group 17 element in a binary compound is -1.
  • The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.
  • The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

Now we have to determine the oxidation state of the elements in the compound.

(a) H_2SO_4

Let the oxidation state of 'S' be, 'x'

2(+1)+x+4(-2)=0\\\\x=+6

Hence, the oxidation state of 'S' is, (+6)

(b) Ca(OH)_2

Let the oxidation state of 'Ca' be, 'x'

x+2(-2+1)=0\\\\x=+2

Hence, the oxidation state of 'Ca' is, (+2)

(c) BrOH

Let the oxidation state of 'Br' be, 'x'

x+(-2)+1=0\\\\x=+1

Hence, the oxidation state of 'Br' is, (+1)

(d) ClNO_2

Let the oxidation state of 'N' be, 'x'

-1+x+2(-2)=0\\\\x=+5

Hence, the oxidation state of 'N' is, (+5)

(e) TiCl_4

Let the oxidation state of 'Ti' be, 'x'

x+4(-1)=0\\\\x=+4

Hence, the oxidation state of 'Ti' is, (+4)

(f) NaH

Let the oxidation state of 'Na' be, 'x'

x+(-1)=0\\\\x=+1

Hence, the oxidation state of 'Na' is, (+1)

4 0
3 years ago
А<br> Newton's Third Law of Motion<br> states:<br> Edit<br> Move with text
Kobotan [32]
Newtown third law of motion says : when two objects interact , the forces they exert on each other are equal and opposite .
4 0
3 years ago
Read 2 more answers
How can ionic bonds and naming be used in a real life example ?
Stells [14]

Answer:

Melting snow more efficiently in winters, understanding the components of mineral water

Explanation:

Let's split this question into two parts. First of all, ionic bonds:

  • an example would be the application of the freezing point depression law. Remember that adding a solute to a specific solvent would decrease the freezing point of a solvent. This is the reason why we add ionic salts, NaCl, to snow in order to make it melt. Knowledge of the fact that 1 mol of NaCl, an ionic compound, dissociates into 2 mol of ions, sodium and chloride, yields us a van 't Hoff factor of 2 rather than 1 for non-electrolytes, molecular compounds. This means the same molality of ionic compounds would produce a twice larger decrease in the freezing point of a solvent;
  • an example for ionic naming is more trivial. Remember the difference between, say, calcium and calcium cation. Sometimes we may read that mineral water is full of calcium. Having chemical knowledge of ionic compound naming would lead us to a conclusion that this is wrong! Mineral water doesn't have any calcium in it, we don't see any metal in mineral water. However, mineral water contains calcium cations, Ca^{2+} and not Ca.
8 0
3 years ago
Calculate the standard molar enthalpy of formation for Na2O(s), given that the standard enthalpy of formation for Na2O2(s) is -5
Anarel [89]

Answer:

- 416 kJ/mol

Explanation:

The standard enthalpy of the reaction (Δ H ∘ rxn) is independent of the pathway, so it can be calculated by the enthalpy of formation of the reactants and the products:

Δ H ∘ rxn = ∑n*Δ H ∘f products - ∑n*Δ H ∘f reactants

Where n is the number of moles in the balanced reaction. So, for the reaction given:

Na₂O(s) + 1/2O₂(g) → Na₂O₂(s)

Because O₂ is formed by only one elements, its Δ H ∘f is 0 kJ/mol:

-89.0 = (1*(-505) - (1*Δ H ∘fNa₂O)

Δ H ∘fNa₂O = -505 + 89

Δ H ∘fNa₂O = - 416 kJ/mol

3 0
4 years ago
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