AN atoms nucleus contains it's protons and electrons true or false

QUESTION 3 What is different between an ionic bond and a covalent bond? An ionic bond is a bond between charged atoms, while a c

Alex73 [517]

Q) What is different between an ionic bond and a covalent bond?

A) An ionic bond is formed between a metal and a non-metal. ... Covalent bonding is a form of chemical bonding between two non metallic atoms which is characterized by the sharing of pairs of electrons between atoms and other covalent bonds.

8 0

3 years ago

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A reaction was performed in which 3.6 g 3.6 g of benzoic acid was reacted with excess methanol to make 1.3 g 1.3 g of methyl ben

Anit [1.1K]

Answer: The percent yield of the reaction is 32.34 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For benzoic acid:

Given mass of benzoic acid = 3.6 g

Molar mass of benzoic acid = 122.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of benzoic acid}=\frac{3.6g}{122.12g/mol}=0.0295mol$

The chemical equation for the reaction of benzoic acid and methanol is:

$\text{Benzoic acid + methanol}\rightarrow \text{methyl benzoate}$

By Stoichiometry of the reaction

1 mole of benzoic acid produces 1 mole of methyl benzoate

So, 0.0295 moles of benzoic acid will produce = $\frac{1}{1}\times 0.0295=0.108$ moles of methyl benzoate

Now, calculating the mass of methyl benzoate from equation 1, we get:

Molar mass of methyl benzoate = 136.15 g/mol

Moles of methyl benzoate = 0.0295 moles

Putting values in equation 1, we get:

$0.0295mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0295mol\times 136.15g/mol)=4.02g$

To calculate the percentage yield of methyl benzoate, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of methyl benzoate = 1.3 g

Theoretical yield of methyl benzoate = 4.02 g

Putting values in above equation, we get:

$\%\text{ yield of methyl benzoate}=\frac{1.3g}{4.02g}\times 100\\\\\% \text{yield of methyl benzoate}=32.34\%$

Hence, the percent yield of the reaction is 32.34 %

6 0

3 years ago

If a 250.0 mL sealed plastic bag starts out at a temperature of 19.0 °C and after sitting in the sun reaches a temperature of 60

Marizza181 [45]

Answer:

D

Explanation:

If the pressure remains constant then the temperature and Volume are all that you have to consider.

Givens

T1 = 19oC = 19 + 273 = 292o K

T2 = 60oC = 60 + 273 = 333oK

V1 = 250 mL

V2 = x

Formula

V1/T1 = V2/T2

250/292 = x/333

Solution.

The solves rather neatly. Multiplly both sides by 333

250*333 / 292 = 333 *x / 333

Do the multiplication

250 * 333 / 292 = x

83250 / 292 = x

Divide by 292

x = 285.1 mL

The answer is D

8 0

2 years ago

A solution is made by combining 500 mL of 0.10 M HF (Ka=7.2 x 10^-4) with 300 mL of 0.15 M NaF. What is the pH of the resulting

n200080 [17]

Answer:

b) 3.10

Explanation:

HF ⇄ H + + F

Using Henderson-Hasselbalch Equation:

pH = pKa + log [A-]/[HA].

Where;

pKa = Dissociation constant = -log Ka

Hence, pKa of HF = -log 7.2 x 10^-4 = 3.14266

[A-] = concentration of conjugate base after dissociation = moles of base/total volume

= 0.15 x 0.3/0.8

= 0.05625 M

[HA] = concentration of the acid = moles of acid/total volume

= 0.10 x 0.5/0.8

= 0.0625 M

Note: Total volume = 500 + 300 = 800 mL = 0.8 dm3

pH = 3.14266 + log [0.05625/0.0625]

= 3.14267 + (-0.04575749056)

= 3.09691250944

From all the available options below:

a) 2.97

b) 3.10

c) 3.19

d) 3.22

e) 3.32

The correct option is b.

4 0

2 years ago

What is the precipitation reaction of sodium nitrate and lead (II)?

Vadim26 [7]

The precipitation reaction of sodium nitrate and lead (II) chloride would not happen. No reaction will happen between these substances. Hope this answers the question. Have a nice day.

8 0

3 years ago