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tia_tia [17]
3 years ago
13

Concentrated hydrochloric acid dilute sulfuric acid chemical equation?

Chemistry
1 answer:
ankoles [38]3 years ago
5 0
H2SO4 (1) H20 (g) + SO3 (g)
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True or false. When an area within an atom is crowded with electrons it is known as a dipole "moment False True
Rama09 [41]

Answer:true

Explanation:

4 0
3 years ago
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What is the mass, in grams, of 0.125 L of CO2 at STP? Question 4 options: 2.80 g 181 g 0.246 g 4.11 g
Zolol [24]
1) Use the fact that 1 mol of gas at STP occupies 22.4 liter

=> 1 mol / 22.4 l = x / 0.125 l => x = 0.125 l * 1 mol / 22.4 l = 0.00558 mol

2) Now use the molar mass of the gas

molar mass of CO2 ≈ 44 g / mol

Formula: molar mass = mass in grams / number of moles =>

mass in grams = molar mass * number of moles = 44 g/mol * 0.00558 moles

mass = 0.246 g

Answer: 0.246 g
7 0
3 years ago
A 0.539 g sample of a metal, M, reacts completely with sulfuric acid according to the reaction M(s)+H2SO4(aq)⟶MSO4(aq)+H2(g) A v
Charra [1.4K]

Answer:

The molar mass of the metal is 54.9 g/mol.

Explanation:

When we work with gases collected over water, the total pressure (atmospheric pressure) is equal to the sum of the vapor pressure of water and the pressure of the gas.

Patm = Pwater + PH₂

PH₂ = Patm - Pwater = 1.0079 bar - 0.03167 bar = 0.9762 bar

The pressure of H₂ is:

0.9762bar.\frac{1atm}{1.013bar} =0.9637atm

The absolute temperature is:

K = °C + 273 = 25°C + 273 = 298 K

We can calculate the moles of H₂ using the ideal gas equation.

P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{0.9637atm \times 0.249L }{(0.08206atm.L/mol.K)\times298K} =9.81 \times 10^{-3} mol

Let's consider the following balanced equation.

M(s) + H₂SO₄(aq) ⟶ MSO₄(aq) + H₂(g)

The molar ratio of M:H₂ is 1:1. So, 9.81  × 10⁻³ moles of M reacted. The molar mass of the metal is:

\frac{0.539g}{9.81 \times 10^{-3} mol} =54.9g/mol

4 0
4 years ago
A mixture containing 20 mole % butane, 35 mole % pentane and rest
notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB

Whereas y accounts for the fractions at the outlet distillate and x for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

0.9=\frac{y_bD}{z_bF}

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

y_bD=0.9*z_bF=0.9*0.2*100mol=18mol

The total distillate flow:

y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol

And the total bottoms flow:

F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025

Then, the molar fraction of pentane and hexane:

x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422

x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

8 0
3 years ago
Read 2 more answers
You have been asked to prove that the Oil Spill Eater speeds up the reaction as you found above. Describe the basic set up of a
andriy [413]

Answer:

The correct answer is option 3. Run a test reaction of crude oil with ocean water over time with Oil Spill Eater present

Explanation:

In any laboratory experiment, all the apparatus needed to carry out a particular experiment must be provided. In this case, our apparatus will be crude oil with ocean water and oil spill eater which is the enzyme used.

We can then run a test reaction of crude oil with ocean water over time with Oil Spill Eater present.

4 0
4 years ago
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