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I need to find the answer for
_ ML = 8,000,000 L

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Answer 1

Answer:

Answer:

To find the value of the blank in the equation "_ ML = 8,000,000 L," you can simply fill in the blank with a number:

"_ ML = 8,000,000 L" can be filled in as "8,000 ML = 8,000,000 L."

So, the answer is 8,000 ML.

Culled from AI

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If 4.0 mol aluminum and 7.0 mol hydrogen bromide react according to the following equation, how many moles of hydrogen are formed and what is the limiting reactant?

Please it's due today

Answers

Answer:

Explanation:

Newton's third law. states that:

Action and reaction are equal and opposite.

I need help!! ASAP please..

Answers

Answer:

164 g

Explanation:

What is the volumetric flow rate in L/s of a stream of air (density = 1 kg/m3) at 1 kg/s?

Answers

Answer:

Volumetric flow rate: Q = 1000 L/s

Explanation:

Volumetric flow rate, also called the rate of fluid flow, is described as volume of fluid that passes a particular point per unit time. The SI unit of volumetric flow rate is m³/s.

Whereas, mass flow rate is defined as the mass of substance that passes through a point per unit of time. SI unit is kg/s.

Given- mass flow rate: ṁ = 1 kg/s and density: ρ = 1 kg/m³

Therefore, volumetric flow rate can be calculated by

$Q = \frac{\dot{m}}{\rho } = (1 kg/s)/(1 kg/m^(3)) = 1 m^(3)/s$

Since, 1 m³/s = 1000 L/s

Therefore, volumetric flow rate: Q = 1 m³/s = 1000 L/s

1. Based on the appearance of your reaction in the beaker, which reagent do you think was consumed and which reagent had some left over? The aluminum was consumed, and copper was left over as seen by the reddish particles. 2. If 5.0 g of iron metal is reacted with 15.0 g of Cl2 gas, how many grams of ferric chloride will form? About 14.52 grams will form. 3. For the reaction above the amount of ferric chloride obtained in the lab was 9.15 grams. Calculate the percent yield. The percent yield would be around 63.02%. 4. What are some reasons for obtaining a percent yield of less than 100 percent? Factors such as the reactants not reacting completely, human error in the experiment, the reactants might have too large of a surface area for reaction, multiple reactions occurring within an experiment, temperature, etc.

Answers

Answer:

1. Al is consumed first and CuSO₄ remains left.

2. The grams of ferric chloride that forms is 14.5 g.

3. The percent yield is 63.1%

4. Trace of impurities present in the reagents, bad manipulations when preparing solutions, etc.

Explanation:

1. The reaction is:

2Al + 3CuSO₄ = Al₂(SO₄)₃ + 3Cu

The number of moles of Al is less than the number of moles of CuSO₄. Therefore, Al is the limiting reagent and CuSO₄ is the excess reagent. This means that Al is consumed first and CuSO₄ remains left.

2. The reaction is:

2Fe + 3Cl₂ = 2FeCl₃

The number of moles of Fe is:

$n_(Fe) =(m_(Fe) )/(MW_(Fe) ) =(5)/(55.85) =0.0895moles$

The number of moles of Cl₂ is:

$n_(Cl2) =(15)/(70.9) =0.211moles$

We know that 2 moles of Fe react with 3 moles of Cl₂, thus:

2 moles Fe---------------3 moles Cl₂

0.0895 moles Fe-------X moles Cl₂

Clearing X:

$Xmoles_(Cl2) =(3*0.0895)/(2) =0.134moles$

It needs 0.134 moles of Cl₂ but it only has 0.211 moles, thus, Cl₂ is the excess reagent. Fe is the limiting reagent.

2 moles Fe-----------2 moles FeCl₃

0.0895 moles Fe------X moles FeCl₃

Clearing X:

$Xmoles_(FeCl3) =(2*0.0895)/(2) =0.0895moles$

$m_(FeCl3) =0.0895molesFeCl3*(162.2gFeCl3)/(1molFeCl3) =14.5g$

3. The actual yield of FeCl₃ is 9.15 g, the theoritical yield is 14.5 g, thus, ther percent yield is:

$Percent-yield=(Actual-yield)/(Theoritical-yield) *100=(9.15)/(14.5) *100=63.1$ %

4. Trace of impurities present in the reagents, bad manipulations when preparing solutions, etc.

Be sure to answer all parts. Industrially, hydrogen gas can be prepared by combining propane gas (c3h8) with steam at about 400°c. The products are carbon monoxide (co) and hydrogen gas (h2). (a) write a balanced equation for the reaction. Include phase abbreviations. (b) how many kilograms of h2 can be obtained from 8.31 × 103 kg of propane

Answers

Balanced chemical reaction:

C₃H₈(g) + 3H₂O(g) → 3CO(g) + 7H₂(g).

M(C₃H₈) = 44.1 g/mol; molar mass of propane.

M(H₂) = 2 g/mol; molar mass of hydrogen.

From balanced chemical reaction: n(C₃H₈) : n(H₂) = 1 : 7.

7m(C₃H₈) : M(C₃H₈) = m(H₂) : M(H₂).

7·8310 kg : 44.1 g/mol = m(H₂) : 2 g/mol.

m(H₂) = 2638.09 kg; mass of hydrogen.

Answer: a) $C_3H_8(g)+3H_2O(g)\rightarrow 3CO(g)+7H_2(g)$

b) $2.64* 10^3kg$

Explanation:

a) According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

$C_3H_8(g)+3H_2O(g)\rightarrow 3CO(g)+7H_2(g)$

b) $\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of propane}=(8.31* 10^6g)/(44.1g/mol)=0.188* 10^6moles$

$C_3H_8(g)+3H_2O(g)\rightarrow 3CO(g)+7H_2(g)$

According to stoichiometry:

1 mole of $C_3H_8$ gives 7 moles of $H_2$

Thus $0.188* 10^6moles$ moles of $C_3H_8$ will give = $(7)/(1)* 0.188* 10^6=1.32* 10^6moles$ of $H_2$

Mass of $H_2=moles* {\text {molar Mass}}=1.32* 10^6moles* 2g/mol=2.64* 10^6g=2.64* 10^3kg$

Thus $2.64* 10^3kg$ of $H_2$ can be obtained from $8.31* 10^3$ kg of propane

Dry ice is solid carbon dioxide. Instead of melting, solid carbon dioxide sublimes according to the equation: CO2(s)→CO2(g). When dry ice is added to warm water, heat from the water causes the dry ice to sublime more quickly. The evaporating carbon dioxide produces a dense fog often used to create special effects. In a simple dry ice fog machine, dry ice is added to warm water in a Styrofoam cooler. The dry ice produces fog until it evaporates away, or until the water gets too cold to sublime the dry ice quickly enough. Suppose that a small Styrofoam cooler holds 15.0 L of water heated to 90 ∘C. Calculate the mass of dry ice that should be added to the water so that the dry ice completely sublimes away when the water reaches 28 ?

Answers

Answer:

6.82 kg

Explanation:

Given that the amount of water is 15L and we know that the density of water is ≈ 1kg/L. The mass of water is given by mass = volume x density, i.e,

mass = 15 x 1 = 15 kg. Also the specific heat capacity of water is 4.186 KJ/kg.

The sublimation enthalpy of dry ice is 571 KJ/kg.

Now, the amount of heat lost by water is entirely used up for the sublimation (conversion from soild to gas) of dry ice. And the heat (Q) lost by water is given as : Q = mCΔT, where m is the mass of water, C the specific heat capacity of water and ΔT the change in temperature.

Here, Q = 15 x 4.186 x (90 - 28) = 3892.98 KJ.

This amount of heat is taken up by the dry ice for its sublimation. Also the energy taken by dry ice (Q') for its sublimation is given by: Q' = m'L', where m' is the mass of dry ice, L' is the latent heat of sublimation (i.e, the amount of heat required per kg of a substance to sublime) of dry ice amd L' = 571 KJ/kg.

Now, Q' =m'L' = heat lost by water = 3892.98KJ.

And, m'L' = m' x 571 KJ/kg = 3892.98 KJ. (Dividing with 571)

Therefore, m' = 6.82 kg.

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