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1 year ago

Does the tempurture of the vinegar affect the volcano flow

1 answer:

6 0

The temperature of vinegar does increase the rate of reaction according to the collision theory.

Hope this answers your question!

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#1: Which type of substance is an electron-pair donor?

Aneli [31]

Answer is: D. Lewis base.
Lewis acid is a chemical element, molecule or ion that contains an empty orbital which is capable of accepting an electron pair from a Lewis base. A Lewis base is element, molecule or ion that has a filled orbital and has electron pair which is not involved in bonding and can give it to Lewis acid. Lewis base is for example ammonia (NH3).

7 0

2 years ago

Never heard of this!!

Ksivusya [100]

Answer:

See explanation

Explanation:

a) u= 60 m/s, v= 45 m/s, t = 7 secs

From;

v = u + at

a = v - u/t

a= 45 - 60/7

a = -2.14 m/s^2

u = 200 m/s

v= 235 m/s

t = 30 secs

v = u + at

a = v-u/t

a = 235 - 200/30

a= 1.17 m/s^2

i) Acceleration = 3 - 1/2-0 = 2/2 = 1 m/s^2

ii) The slope between 2 and 5 is zero because velocity was not changing with time.

iii) Within the time between 2 and 5 the speed remained at 3 m/s

iv) the slope between 5 and 8 is 2 - 3/8 - 5 = -1/3 = -0.33m/s^2

v) The negative slope means that velocity is decreasing with time(deceleration).

7 0

2 years ago

Question 4

vazorg [7]

Answer:

H(aq) + NO3 (aq) + HF(aq)

Explanation:

In the given mixture of HNO3 (Nitric Acid) and HF (hydrofluoric acid) in water the major species present are H(aq) + NO3 (aq) + HF(aq).

On the reaction of HNO3 (Nitric Acid) and HF (hydrofluoric acid) in water , it will give a polar solution and will form a homogenous mixture.

Hence, the correct answer is "H(aq) + NO3 (aq) + HF(aq)".

4 0

2 years ago

The molar mass of a certain gas is 80.6 g. What is the density of the gas in g/L at STP?

vfiekz [6]

Using the ideal gas equation:
PV = nRT

Substituting n with mass / Mr
PV = mRT/Mr

Density = m/V
So rearranging:
Density = PMr/RT

P = 1 atm
R = 0.082 L atm / K mol
T = 273 K
Density = (1 x 80.6) / (0.082 x 273)
Density = 3.6 g / L

3 0

2 years ago

Heyy guys, so basically i need help with stoichiometric calculation I will give you 100 points just to answer all of these answe

jeka94

Answer:

3. The mass of ethanol required is approximately 0.522869 g

The mass of ethanoic acid required is approximately 0.68156 g

4. The mass of iron (III) oxide required is approximately 285.952.189.095 tonnes

5. The mass of silver nitrate required is approximately 14.53 grams

6. The mass of copper oxide that would be needed is approximately 31.86 grams

7. a. The mass of the precipitate, Zn(OH)₂ formed is approximately 49.712 grams

b. The mass of the precipitate, Al(OH)₃ formed is approximately 13 grams

c. The mass of the precipitate, Mg(OH)₂, formed is approximately 14.579925 grams

Explanation:

3. The 1 mole of ethanol and 1 mole of ethanoic acid combines to form 1 mole of ethyl ethanoate

The number of moles of ethyl ethanoate in 1 gram of ethyl ethanoate, n = 1 g/(88.11 g/mol) = 1/88.11 moles

∴ The number of moles of ethanol = 1/88.11 moles

The number of moles of ethanoic acid = 1/88.11 moles

The mass of ethanol = (46.07 g/mol) × 1/88.11 moles = 0.522869 g

The mass of ethanoic acid in the reaction = 60.052 g/mol × 1/88.11 moles ≈ 0.68156 g

4. 1 mole of iron(III) oxide reacts with 1 mole of CO₂ to produce 1 mole of iron

The number of moles in 100 tonnes of iron= 100000000/55.845 = 1790670.60614 moles

The mass of iron (III) oxide required = 159.69 × 1790670.60614 = 285952189.095 g ≈ 285.952.189.095 tonnes

5. The number of moles of NaCl in 5 grams of NaCl = 5 g/58.44 g/mol = 0.0855578371 moles

The mass of silver nitrate required, m = 169.87 g/mol × 0.0855578371 moles ≈ 14.53 grams

6. The number of moles of CuSO₄·5H₂O in 100 g of CuSO₄·5H₂O = 100 g/(249.69 g/mol) ≈ 0.4005 moles

The mass of copper oxide required, m = 79.545 g/mol × 0.4005 moles ≈ 31.86 grams

7. a. The number of moles of NaOH in the reaction = 20 g/(39.997 g/mol) ≈ 0.5 moles

2 moles of NaOH produces 1 mole of Zn(OH)₂

0.5 moles of NaOH will produce 0.5 mole of Zn(OH)₂

The mass of 0.5 mole of Zn(OH)₂ = 0.5 mole × 99.424 g/mol = 49.712 grams

The mass of the precipitate, Zn(OH)₂ formed = 49.712 grams

b. 6 moles of NaOH produces 2 moles Al(OH)₃

20 g, or 0.5 mole of NaOH will produce (1/6) mole of Al(OH)₃

The mass of the precipitate, Al(OH)₃ formed, m = 78 g/mol×(1/6) moles = 13 grams

c. 2 moles of NaOH produces 1 mole of Mg(OH)₂, therefore;

20 g or 0.5 moles of NaOH formed (1/4) mole of Mg(OH)₂

The mass of the precipitate, Mg(OH)₂, formed, m = 58.3197 g/mol × (1/4) moles = 14.579925 grams

3 0

2 years ago

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