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rewona [7]
4 years ago
8

Which substance has a high melting point and conducts electricity in the liquid phase

Chemistry
2 answers:
ELEN [110]4 years ago
8 0
It is true that substances that have a high melting point and conduct electricity in the liquid phase are ionic substances, and if one of your options is NaCl, then that is the correct answer.
cluponka [151]4 years ago
3 0
Substances that are ionic compounds
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Where was all the hydrogen in the universe formed?
Maslowich
The low-mass elements, hydrogen and helium, were produced in the hot, dense conditions of the birth of the universe itself. The birth, life, and death of a star is described in terms of nuclear reactions. The chemical elements that make up the matter we observe throughout the universe were created in these reactions.
5 0
3 years ago
Consider the following balanced redox reaction:2CrO₂⁻(aq) + 2H₂O(l) + 6ClO⁻(aq) →
LUCKY_DIMON [66]

The balanced  redox reaction are given below

2CrO₂⁻(aq) + 2H₂O(l) + 6ClO⁻(aq) →2CrO₄²⁻(aq) + 3Cl₂(g) + 4OH⁻(aq)

The Cl species is oxidizing agent .

There are following steps involve in balanced redox reaction in basic medium .

  • Divide the reaction into two half reactions .
  • Balance the elements other than oxygen and hydrogen .
  • Balance the oxygen atom by adding H_{2} O  .
  • Balance the hydrogen atom by adding H^{+} .
  • Balance the charge  by adding e^{-} .
  • Add OH^{-} on both side
  • Neutralize by combing  OH^{-} and  H^{+} .
  • Add the half reaction and simply .

The species which reduces the other species in the redox reaction is called oxidizing agent . Similarly , the species which oxidize the other species in the redox reaction  is called reducing agent .

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7 0
1 year ago
Which state of matter would demonstrate the best ability to flow/pour?explain using the term particles
frutty [35]

Liquid, in a liquid state of matter, particles will flow over eachother, the forces between the particles are strong enough to hold a certain volume but not strong enough to keep the molecules sliding over each other.


4 0
3 years ago
If you needed to make 100 mL of a 0.2 M fruit drink solution from the 1.0 M fruit drink solution, how would you do it? (Hint: Us
Studentka2010 [4]

Answer: We take 20 ml of 1.0 M fruit drink and add 80 ml of water to get 100 ml of 0.2 M solution.

Explanation:

According to the neutralization law,

M_1V_1=M_2V_2

where,

M_1 = molarity of stock solution = 0.2 M

V_1 = volume of stock solution = 100 ml

M_2 = molarity of resulting solution = 1.0 M

V_2 = volume of resulting solution = ?

Now put all the given values in the above law, we get the volume of resulting solution.

(0.2M)\times 100=(1.0M)\times (V_2)

V_2=20ml

Therefore, the volume of 1.0 M required is 20 ml.

We take 20 ml of 1.0 M fruit drink and add 80 ml of water to get 100 ml of 0.2 M solution.

7 0
3 years ago
Calculate the volume in ) of 0.100 M Na2CO3 needed to produce 1.00 g of CaCO 3 (s) . There is an excess of CaCl 2. What’s the vo
luda_lava [24]

Answer:

100 mL of Na2CO3

Explanation:

We'll begin by calculating the number of mole in 1 g of CaCO3. This can be obtained as follow:

Mass of CaCO3 = 1 g

Molar mass of CaCO3 = 100.09 g/mol

Mole of CaCO3 =?

Mole = mass /Molar mass

Mole of CaCO3 = 1/100.09

Mole of CaCO3 = 0.01 mole

Next, we shall determine the number of mole of Na2CO3 needed to produce 0.01 mole of CaCO3.

This is illustrated below:

Na2CO3 + CaCl2 —> 2NaCl + CaCO3

From the balanced equation above,

1 mole of Na2CO3 reacted to produce 1 mole of CaCO3.

Therefore, 0.01 mole of Na2CO3 will also react to produce 0.01 mole of CaCO3.

Next, we shall determine the volume of Na2CO3 needed for the reaction as illustrated below:

Mole of Na2CO3 = 0.01 mole

Molarity of Na2CO3 = 0.1 M

Volume of Na2CO3 solution needed =?

Molarity = mole /Volume

0.1 = 0.01 / volume of Na2CO3

Cross multiply

0.1 × volume of Na2CO3 = 0.01

Divide both side by 0.1

Volume of Na2CO3 = 0.01 / 0.1

Volume of Na2CO3 = 0.1 L

Finally, we shall convert 0.1 L to millilitres (mL). This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.1 L = 0.1 L × 1000 mL / 1 L

0.1 L = 100 mL

Thus, 0.1 L is equivalent to 100 mL.

Therefore, 100 mL of Na2CO3 is needed for the reaction.

5 0
3 years ago
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