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WINSTONCH [101]
3 years ago
6

Chlorine and potassium atoms form ionic bonds, carbon atoms form non-polar covalent bonds with nitrogen atoms, and oxygen forms

polar covalent bonds with phosphorus. Explain why these bonds are the types they are.
Chemistry
1 answer:
Ray Of Light [21]3 years ago
5 0

Explanation:

A covalent bond is formed when an element shares its valence electron with another element. This bond is formed between two non metals.

An ionic bond is formed when an element completely transfers its valence electron to another element. The element which donates the electron is known as electropositive element and the element which accepts the electrons is known as electronegative element. This bond is formed between a metal and an non-metal.

Chlorine and potassium atoms form ionic bonds: Ionic bond is formed when there is complete transfer of electron from a highly electropositive metal to a highly electronegative non metal.  Electronegativity difference = electronegativity of chlorine - electronegativity of potassium = 3-0.8 = 2.2

Carbon atoms form non-polar covalent bonds with nitrogen atoms : Non-polar covalent bond is defined as the bond which is formed when there is no difference of electronegativities between the atoms.  Electronegativity difference  = electronegativity of nitrogen  - electronegativity of carbon= 3.0-2.5 = 0.5

Oxygen forms polar covalent bonds with phosphorus: A polar covalent bond is defined as the bond which is formed when there is a difference of electronegativities between the atoms.  Electronegativity difference = electronegativity of oxygen - electronegativity of phosphorous = 3.5- 2.19 = 1.31

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. In which type of ecosystem would fire most likely be a limiting factor for plant species?
bulgar [2K]

Answer:

A forest or woodland area

Explanation:

Forests or woodland areas have lots of plants and grass like things that wuld burn easily

8 0
3 years ago
A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of
Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the <u>reaction</u> between HCl and KOH:

HCl~+~KOH->~H_2O~+~KCl

Now <u>for example, we can use a volume of 10 mL of HCl</u>. So, we can calculate the moles using the <u>molarity equation</u>:

M=\frac{mol}{L}

We know that 10mL=0.01L and we have the concentration of the HCl 0.282M, when we plug the values into the equation we got:

0.282M=\frac{mol}{0.01L}

mol=0.282*0.01

mol=0.00282

We can do the same for the KOH values (50mL=0.05L and 0.141M).

0.141M=\frac{mol}{0.05L}

mol=0.141*0.05

mol=0.00705

So, we have so far <u>0.00282 mol of HCl</u> and <u>0.00705 mol of KOH</u>. If we check the reaction we have a <u>molar ratio 1:1</u>, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an <u>excess of KOH</u>. This excess can be calculated if we <u>substract</u> the amount of moles:

0.00705-0.00282=0.00423mol~of~KOH

Now, if we want to calculate the pH value we will need a <u>concentration</u>, in this case KOH is in excess, so we have to calculate the <u>concentration of KOH</u>. For this, we already have the moles of KOH that remains left, now we need the <u>total volume</u>:

Total~volume=50mL+10mL=60mL

60mL=0.06L

Now we can calculate the concentration:

M=\frac{0.00423mol}{0.06L}

M=0.0705

Now, we can <u>calculate the pOH</u> (to calculate the pH), so:

pOH=-Log(0.0705)

pOH=1.15

Now we can <u>calculate the pH value</u>:

14=~pH~+~pOH

pH=14-1.15=12.84

8 0
3 years ago
The decomposition reaction of N2O5 in carbon tetrachloride is 2N2O5−→−4NO2+O2. The rate law is first order in N2O5. At 64 °C the
Mademuasel [1]

Answer:

(a) rate =  4.82 x 10⁻³s⁻¹  [N2O5]

(b) rate =   1.16 x 10⁻⁴  M/s

(c) rate =   2.32 x 10⁻⁴ M/s

(d) rate =   5.80 x 10⁻⁵ M/s

Explanation:  

We are told the rate law is first order in N₂O₅, and its rate constant is 4.82 x 10⁻³s⁻¹ . This means the rate is proportional to the molar concentration   of   N₂O₅, so

(a) rate = k [N2O5] = 4.82 x 10⁻³s⁻¹ x [N2O5]

(b) rate = 4.82×10⁻³s⁻¹  x 0.0240 M =  1.16 x 10⁻⁴ M/s

(c) Since the reaction is first order if the concentration of  N₂O₅ is double the rate will double too:  2 x   1.16 x 10⁻⁴ M/s = 2.32 x 10⁻⁴ M/s

(d) Again since the reaction is halved to 0.0120 M, the rate will be halved to

1.16 x 10⁻⁴ M/s / 2 =  5.80 x 10⁻⁵ M/s

3 0
3 years ago
Read 2 more answers
A 2.50-l volume of hydrogen measured at â196 °c is warmed to 100 °c. calculate the volume of the gas at the higher temperature
ivann1987 [24]

To solve this we assume that the hydrogen gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant pressure and number of moles of the gas the ratio T/V is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

T1 / V1 = T2 / V2

V2 = T2 x V1 / T1

V2 = (100 + 273.15) K x 2.50 L / (-196 + 273.15) K

<span>V2 = 12.09 L</span>

Therefore, the volume would increase to 12.09 L as the temperature is increased to 100 degrees Celsius.

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5 0
3 years ago
Based on the balanced equation 2C2H2 + 5O2 → 4CO2 + 2H2O calculate the number of excess reagent units remaining when 52 C2H2 mol
erica [24]

Answer:

20 molecules of oxygen gas remains after the reaction.

Explanation:

2C_2H_2 + 5O_2\rightarrow 4CO_2 + 2H_2O

Molecules of ethyne = 52

Molecules of oxygen gas = 150

According to reaction, 2 molecules of ethyne reacts with 5 molecules of oxygen gas.

Then 52 molecules of ethyne will react with:

\frac{5}{2}\times 52=130 molecules of oxygen gas.

As we can see that we have 150 molecules of oxygen gas, but 52 molecules of ethyne will react with 130 molecules of oxygen gas. So, this means that ethyne is a limiting reagent and oxygen gas is an excessive reagent.

Remaining molecules of recessive reagent = 150 - 130 = 20

20 molecules of oxygen gas remains after the reaction.

5 0
3 years ago
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