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3 years ago

A/1 * B/C = D

Chemistry

1 answer:

grigory [225]3 years ago

3 0

Answer : 4.8 florins can you exchange for your 4 guilders.

Solution : Given,

The current exchange rate for every 2.4 florins is 2.0 guilders

As per question,

The exchange rate is 2.0 guilders for every 2.4 florins

The exchange rate is 1 guilders for every $\frac{2.4}{2.0}$ florins

The exchange rate is 4 guilders for every $\frac{2.4}{2.0}\times 4=4.8$ florins

Therefore, the 4.8 florins can you exchange for your 4 guilders.

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Do you think that participation in a Supervised Agriculture Experience (SAE) prepares a student for a career in the AFN industry

irinina [24]

Answer:

The SAE curriculum includes practical farming tasks conducted outside the scheduled classroom and laboratory period by students. SAEs offer a method for students in agricultural education to gain real-world work opportunities that they are most interested in in the field of agriculture. Supervised agricultural experience is an essential component of agricultural education, and all Agriculture, Food and Natural Resources (AFNR) courses are a necessary component.

Explanation: Hope it helps

6 0

3 years ago

A water treatment plant applies chlorine for disinfection so that 10 mg/L chlorine is achieved immediately after mixing. The vol

Vlada [557]

Answer: The mass of chlorine needed by the plant per day is $9.4625\times 10^8mg$

Explanation:

We are given:

Volume o water treated per day = 25,000,000 gallons

Converting this volume from gallons to liters, we use the conversion factor:

1 gallon = 3.785 L

So, $\frac{3.785L}{1\text{ gallon}}\times 25,000,000\text{ gallons}=9.4625\times 10^7L$

Amount of chlorine applied for disinfection = 10 mg/L

Applying unitary method:

For 1 L of water, the amount of chlorine applied is 10 mg

So, for $9.4625\times 10^7L$ of water, the amount of chlorine applied will be $\frac{10mg}{1L\times 9.4625\times 10^7L}=9.4625\times 10^8mg$

Hence, the mass of chlorine needed by the plant per day is $9.4625\times 10^8mg$

6 0

3 years ago

If 4000 g of Fe2O3 reacts, how many moles of CO are needed? (add work)

7nadin3 [17]

Answer:

75.15 mol.

Explanation:

Firstly, we need to write the balanced equation of the reaction:

Fe₂O₃ + 3CO → 2Fe + 3CO₂.

It is clear that 1.0 mole of Fe₂O₃ reacts with 3.0 moles of CO to produce 2.0 moles of Fe and 3.0 moles of CO₂.

∴ Fe₂O₃ reacts with CO with (1: 3) molar ratio.

we need to calculate the no. of moles of (4000 g) of Fe₂O₃:

no. of moles of Fe₂O₃ = mass/molar mass = (4000 g)/(159.69 g/mol) = 25.05 mol.

Using cross multiplication:

1.0 mole of Fe₂O₃ needs → 3.0 moles of CO,

∴ 25.05 mole of Fe₂O₃ needs → ??? moles of CO.

∴ The no. of moles of CO needed = (3.0 mol)(25.05 mol)/(1.0 mol) = 75.15 mol.

5 0

3 years ago

A rectangular corral of widths Lx = L and Ly = 2L contains seven electrons. What is the energy of (a) the first excited state, (

DiKsa [7]

Answer:

See explaination

Explanation:

The invariant mass of an electron is approximately9. 109×10−31 kilograms, or5. 489×10−4 atomic mass units. On the basis of Einstein's principle of mass–energy equivalence, this mass corresponds to a rest energy of 0.511 MeV.

Check attachment for further solution to the exercise.

3 0

3 years ago

You are requested to reduce the size of 50 ton/hr of a given solid. The size of the feed is such 80% passes a 4-in (76.2 mm) scr

defon

Answer:

1) The power needed to process 50 ton/hr is 135.4 HP.

2) The void fraction of the bed is 0.37.

Explanation:

1) For this type of milling operations, we can estimate the power needed for an operation according to the work index (Ei), the passing size of the circuit feed (F80) and the passing size of the product (P80).

We assume the units of Ei are kWh/t.

The equation that relates this parameters and the power is (size of particles in μm):

$W=Ei*(\frac{10}{\sqrt{P80}} -\frac{10}{\sqrt{F80}} )\\\\W=9.45*(\frac{10}{\sqrt{3175\mu m}} -\frac{10}{\sqrt{76200mm}} )\\\\\\W=9.45*(0.1774+0.0362)=2.019 kWh/t$

The power needed to process 50 ton/hor is

$P=2.0194\frac{kWh}{Ton}*\frac{50Ton}{h}*\frac{1.341HP}{1kW}= 135.4 \, HP$

2) The density of the packed bed can be expressed as

$\rho=f_v*\rho_v+f_s*\rho_s$

being f the fraction and ρ the density of every fraction. We know that the density of the void is 0 (ρv=0) and that fv=1-fs (the sum of the fractions ois equal to the total space).

Then we can rearrange

$\rho=f_v*\rho_v+f_s*\rho_s\\\\\rho=f_v*0+(1-f_v)*\rho_s\\\\\rho/\rho_s=1-f_v\\\\f_v=1-\rho/\rho_s=1-990/1570=1-0.63=0.37$

The void fraction of the bed is 0.37.

6 0

3 years ago