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Illusion [34]
3 years ago
8

Define and explain bond energy.

Chemistry
2 answers:
Tomtit [17]3 years ago
6 0
Bond Energy - measure of the amount of energy needed to break apart one molecule of covalently bonded gases.
julia-pushkina [17]3 years ago
4 0
Bond energy<span> is known as the amount of </span>energy<span> needed to break a molecule into its atoms.  </span><span>.</span>
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please don't search this up thx or ill taker the points back- the question I need help is-What are respiratory disorders? Explai
irinina [24]

These are  diseases that affect primarily the lungs or any other part of the respiratory system.

<h3>What are respiratory disorders?</h3>

The term respiratory disorders refers to those diseases that affect primarily the lungs or any other part of the respiratory system. These are the diseases that prevent a person from breathing in and out normally and prevent the lungs from playing its usual role in gaseous exchange.

The respiratory disorders could be able to make a person not to be able to breathe well and it may lead to hospitalization and death of the patient of adequate treatment is not sought from an expert who could diagnose and treat such respiratory disorders.

The common types of respiratory disorders are;

  • asthma
  • cystic fibrosis
  • emphysema
  • lung cancer
  • mesothelioma
  • tuberculosis

Learn more about respiratory disorder:brainly.com/question/15366562

#SPJ1

8 0
1 year ago
2. Arnie is going to dissolve 2.65g of zinc using 6.0M hydrochloric acid. A) what volume of the acid will he need to dissolve al
fenix001 [56]

Answer:

Explanation:

Zn + 2HCl = ZnCl₂ + H₂

A ) mole of Zn = 2.65 / 65

= .04 mol

mole of HCl required = .04 x 2 mol

.08 mol

If v be the volume required

v x 6 = .08

v = .0133 liter

= 13.3 cc

B )

volume of gas at NTP :

moles of gas  obtained = .04 moles

= 22.4 x .04 liter

= .896 liter

we have to find this volume at given temperature and pressure

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

\frac{760\times.896}{273} =\frac{628\times V_2}{296}

V_2 = 1.175 liter.

C )

.04 mole of zinc chloride will be produced

mol weight of zinc chloride

= 65 + 35.5 x 2

= 136 gm

.04 mole = 136 x .04

= 5.44 gm

5 0
3 years ago
Evaporation and condensation
bagirrra123 [75]

Answer:

Evaporation and condensation are two processes through which matter changes from one state to another.

Explanation:

7 0
4 years ago
Read 2 more answers
If a compound has a composition of 82% nitrogen and 18% hydrogen, what is the empirical formula for this compound
Damm [24]

Answer: The empirical formula for the given compound is NH_3

Explanation : Given,

Percentage of H = 18 %

Percentage of N = 82 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of H = 18 g

Mass of N = 82 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{18g}{1g/mole}=18moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{82g}{14g/mole}=5.8moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 5.8 moles.

For Hydrogen  = \frac{18}{5.8}=3.10\approx 3

For Nitrogen = \frac{5.8}{5.8}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of H : N = 3 : 1

Hence, the empirical formula for the given compound is NH_3

3 0
3 years ago
How many moles of co2 are produced when 5.40 mol of ethane are burned in an excess of oxygen?
Nimfa-mama [501]
<span>Reaction of ethane combustion:

2C2H6 (g) + 7O2 (g) ----> 4CO2 (g) + 6H2O 

According to the reaction, we can see that </span>C2H6 and CO2 have following stoichiometric ratio: 

n(C2H6) : n(CO2) = 2 : 4

If we know the number of moles of ethane we can calculate moles of carbon dioxide:

2 x n(CO2) = 4 x n(C2H6)

n(CO2) = 2 x n(C2H6)

n(CO2) = 2 x 5.4 = 10.8 mole of CO2 <span>are produced</span>


3 0
4 years ago
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