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AnnZ [28]
3 years ago
10

Omg pls help i dunno what the frick frack this is

Chemistry
1 answer:
snow_lady [41]3 years ago
4 0

Answer:

1. Mass of KCl produced = 774.8 g of KCl

2. Mass of KNO₃ produced = 13.837g

3. Moles of NaOH made = 0.846 moles

4. Moles of LiCl produced = 0.846 moles

5. Moles of CO₂ produced = 207.6 moles

Explanation:

1. From the equation of reaction, 1 mole of ZnCl₂ produces, 2 moles of KCl.

5.02 moles of ZnCl₂ will produce, 2 × 5.02 moles of KCl = 10.4 moles of KCl

Molar mass of KCl = (39 + 35.5) g/mol = 74.5 g/mol

10.4 moles of KCl = 10.4 × 74.5 g

Mass of KCl produced = 774.8 g of KCl

2. Mole ratio of KNO₃ and KOH = 1:1

O.137 moles of KOH will produce 0.137 moles of KNO₃

Molar mass of KNO₃ = 101 g/mol

Mass of KNO₃ produced = 0.137 × 101 g = 13.837g

3. Molar mas of Ca(OH)₂ = 74.0 g

Moles of Ca(OH)₂ in 31.3 g = 31.3/74.0 = 0.423 moles of Ca(OH)₂

Mole ratio of NaOH and Ca(OH)₂ in the reaction = 2 : 1

Moles of NaOH made = 2 × 0.423 = 0.846 moles

4. Molar mass of MgCl₂ = 95.0 g

Moles of MgCl₂ in 40.2 g = 40.2/95.0 = 0.423 moles

From the reaction equation, mole ratio of MgCl₂ and LiCl = 1:2

Moles of LiCl produced = 2 × 0.423 = 0.846 moles

5. From the equation of reaction, 1 mole of C₆H₁₀O₅ produces 6 moles of cO₂

34.6 moles of C₆H₁₀O₅ will produce 34.6 × 6 moles of CO₂

Moles of CO₂ produced = 207.6 moles

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Explanation:

6 0
3 years ago
Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 1.30 MJ o
Sunny_sXe [5.5K]

Answer:

97 000 g Na

Explanation:

The absortion (or liberation) of energy in form of heat is expressed by:

q=m*Cp*ΔT

The information we have:

q=1.30MJ= 1.30*10^6 J

ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)

Cp=30.8 J/(K mol Na)

If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.

To do so, we use the molar mass of Na= 22.99g/mol

Cp= \frac{30.8J}{K*mol Na}*\frac{1 mol Na}{22.99 g Na}=1.34\frac{J}{K*g Na}

Now, we are able to solve for m:

m=\frac{1.30*10^6 J}{1.34\frac{J}{K*g Na} *10.0K}=\frac{1.30*10^6J}{13.4\frac{J}{g Na} }  = 9.70*10^4 g Na=97 Kg Na=97 000 g Na

7 0
3 years ago
Read 2 more answers
If the volume and Kelvin temperature are reduced by one-half what will happen to the pressure???
jolli1 [7]

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Explanation:

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3 years ago
1 C8H10(l) +21/2O2(g) → 8CO2(g) + 5H2O(g), Hcomb= ? Hf for C8H10(l) = +49.0kJ/mol C8H10(l) Use the balanced combustion reaction
nikklg [1K]

Answer:

H_{comb}=-4406kJ/mol

Explanation:

Hello,

In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

H_{comb}=8*\Delta _fH_{CO_2}+5\Delta _fH_{H_2O}-\Delta _fH_{C_8H_{10}}

Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:

H_{comb}=8*(-393.5kJ/mol)+5(-241.8kJ/mol)-49.0kJ/mol\\\\H_{comb}=-4406kJ/mol

Best regards.

4 0
3 years ago
Please answer he questions
Nastasia [14]

"Work done by a constant force on an object is the product of the force and the distance moved by the object in the direction of the force" -textbook

There is work done ONLY if the direction of the force and the direction the item is moving are the same. In figure A, the direction of the force (the lifting) is upwards (defying gravity), and the book is moving upwards, so work is done. In figure B, the force is still moving upwards (the person is carrying the books) but the book is moving to the right, so there is NO work done.

6 0
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