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Savatey [412]
3 years ago
7

What's the difference between a conductor and an insulator?I'll mark you as brainliest ​

Chemistry
2 answers:
irina [24]3 years ago
8 0

conductor are defined as the materials or substances that allow electricity to flow through them. Also, conductors allow heat to be transmitted through them. Examples of conductors are metals, the human body, Earth and animals. The human body is a strong conductor.

Any material that keeps energy such as electricity, heat, or cold from easily transferring through is an insulator. Wood, plastic, rubber, and glass are good <em><u>insulator</u></em><em><u>.</u></em>

Explanation:

have a beautiful day

prisoha [69]3 years ago
6 0
Heat and electricity travel via a conductor. Insulator blocks current and heat flow.

The majority of metals are excellent conductors.

Heat-insulated pipes handle hot liquids and gases.

I hope this helps you
:)
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Give the formula of the conjugate base of HSO4–
mr Goodwill [35]
The answer is So^2-4
7 0
3 years ago
What is the ph of a soft drink in which the major buffer ingredients are 6.6 g of nah2po4 and 8.0 g of na2hpo4 per 355 ml of sol
Alex777 [14]
The Relative Formula Mass of NaH2PO4 is 120 g/mol
Therefore, the number of moles = 6.6/120
                                                   = 0.055 moles of NaH2PO4 which is also equal to the number of moles of H2PO4.
[H2PO4-] = Number of moles oof H2PO4-/Volume of the solution in L
  = 0.055/ ( 355 ×10^-3)
  = 0.155 M
Na2HPO4 undergoes complete dissociation as follows;
Na2HPO4 (aq)= 2Na+ (aq) + HPO4^2- (aq)

1 mole of Na2HPO4 = 142 g/mol
Therefore; number of moles = 8.0/142
                                             = 0.0563 moles
 [HPO4 ^-2] is given by no of moles HPO4^2- /volume of the solution in L
     = 0.0563/(355×10^-3)
     =  0.1586 M
Both H2PO4^2- and HPO4^2- are weak acids the undergoes partial dissociation 
Ka of H2PO4- = 6.20 × 10^-8
 [H+] =Ka*([H2PO4-]/[HPO4(2-)]
        = (6.20 ×10^-8)×(0.155/0.1586)
        = 6.059 ×10^-8 M
pH = - log[H+]
     = - log (6.059×10^-8)
     = 7.218

5 0
3 years ago
What Element am I?
rjkz [21]

Answer:

The correct answer is - sulfur.

Explanation:

In the periodic table, there are 18 groups and 7 rows or periods arranged according to their atomic number or electronic configuration. In the question, it is mentioned that the desired element atomic mass is less than the atomic mass of the selenium which is 78.96, and more than oxygen which is 15.99 with 6 electron valence and present in the third row.

As it has 6 valency of electron it must be in the 16 group of the table that comprises the 6 valency and as it is located in the 3rd row it must be sulfur that also has an atomic mass between selenium and oxygen.

6 0
3 years ago
______ reactions release energy.
Igoryamba
Combustion reactions release energy in the form of heat. 
        Hope this helps and tell me if you need more help!
4 0
3 years ago
A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at
lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is  -5485.7 kJ/mol

Explanation:

<u>Step 1:</u> Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

<u> Step 2:</u> Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

<u>Step 3:</u> Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25  = 8.2 °C

q = 1200 * 4.184 * 8.2 =  41170.56 J

<u>Step 4</u>: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J  = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

<u>Step 5</u>: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

<u>Step 6:</u> Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is  -5485.7 kJ/mol

8 0
3 years ago
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