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zmey [24]
3 years ago
8

In normal conditions, warm water "piles up" in the Western Pacific Ocean. True False

Chemistry
1 answer:
kakasveta [241]3 years ago
8 0
In normal conditions, warm water does "pile up" in the" Western Pacific Ocean.
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Which best describes the energy conversion that happens when hydroelectric
Sliva [168]

Answer:

"A hydraulic turbine converts the energy of flowing water into mechanical energy. A hydroelectric generator converts this mechanical energy into electricity.

Explanation:

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4 0
3 years ago
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Which two values for `DeltaG` and E°cell correctly indicate a spontaneous reaction?
Mila [183]

Well as an spontaneous reaction will have a negative gibbs free energy (delta g) and a positive e cell and also the E cell is also measured in V, then your answer is the first one: ΔG = –295kJ, E°cell = +1.53 v. Hope this works



4 0
3 years ago
Is NaHCO3 a compound? Explain pls!
Slav-nsk [51]

Answer:

Yes it is

Explanation: because it is composed of identical molecules consisting of atoms of two or more chemical elements.

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4 0
2 years ago
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Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N
Gnesinka [82]

Answer:

NH3 is the limiting reactant

The theoretical yield is 216.0 kg urea

The % for this reaction is 78.8 %

Explanation:

<u>Step 1:</u> Data given

Mass of ammonia = 122.5 kg

Mass of carbon dioxide = 211.4 kg

Mass of urea produced = 170.3 kg

Molar mass of ammnoia = 17.031 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Moalr mass of urea = 60.06 g/mol

<u>Step 2:</u> The balanced equation

2NH3(aq) + CO2(aq) --> CH4N2O(aq) + H2O(l)

<u>Step 3:</u> Calculate moles of NH3

Number of moles = mass / Molar mass

Moles NH3 = 122500 grams / 17.031 g/mol

Moles NH3 = 7192.77 moles

<u>Step 4:</u> Calculate moles of CO2

Moles CO2 = 211400 / 44.01 g/mol

Moles CO2 = 4803.45 moles

<u>Step 5</u>: Calculate limiting reactant

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

NH3 is the limiting reactant. It will completely be consumed (7192.77 moles).

CO2 is in excess. There will be consumed 7192.77/2 = 3596.4 moles

There will remain 4803.45 - 3596.4 = 1207.05 moles of CO2

<u>Step 6:</u> Calculate moles of urea produced:

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

For 7192.77 moles of NH3, we have 3596.4 moles of urea produced

<u>Step 7: </u>Calculate mass of urea

Mass urea = moles urea * molar mass urea

Mass urea = 3596.4 moles * 60.06 g/mol

Mass urea = 216000 grams = 216 kg = theoretical yield

<u>Step 8</u>: Calculate % yield

% yield = (actual yield / theoretical yield)*100%

% yield = (170.3 / 216) *100% = 78.8%

The % for this reaction is 78.8 %

3 0
3 years ago
The reaction of equal molar amounts of benzene, C6H6, and chlorine, Cl2, carried out under special conditions, yields a gas and
WINSTONCH [101]

Answer : The molecular formula of a compound is, C_6H_5Cl_1

Solution : Given,

Mass of C = 64.03 g

Mass of H = 4.48 g

Mass of Cl = 31.49 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of Cl = 35.5 g/mole

Step 1 : convert given masses into moles.

Moles of C = \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{64.03g}{12g/mole}=5.34moles

Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{4.48g}{1g/mole}=4.48moles

Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{31.49g}{35.5g/mole}=0.887moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{5.34}{0.887}=6.02\approx 6

For H = \frac{4.48}{0.887}=5.05\approx 5

For Cl = \frac{0.887}{0.887}=1

The ratio of C : H : Cl = 6 : 5 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = C_6H_5Cl_1

The empirical formula weight = 6(12) + 5(1) + 1(35.5) = 112.5 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}

n=\frac{112.5}{112.5}=1

Molecular formula = (C_6H_5Cl_1)_n=(C_6H_5Cl_1)_1=C_6H_5Cl_1

Therefore, the molecular of the compound is, C_6H_5Cl_1

5 0
3 years ago
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