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2 years ago

6K + B2O3 → 3K2O + 2B

Chemistry

1 answer:

atroni [7]2 years ago

6 0

Answer:

104.84 moles

Explanation:

Given data:

Moles of Boron produced = ?

Mass of B₂O₃ = 3650 g

Solution:

Chemical equation:

6K + B₂O₃ → 3K₂O + 2B

Number of moles of B₂O₃:

Number of moles = mass/ molar mass

Number of moles = 3650 g/ 69.63 g/mol

Number of moles = 52.42 mol

Now we will compare the moles of B₂O₃ with B from balance chemical equation:

B₂O₃ : B

1 : 2

52.42 : 2×52.42 = 104.84

Thus from 3650 g of B₂O₃ 104.84 moles of boron will produced.

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Answer: The specific heat of metal is 2.34 J/g°C

Explanation:

To calculate the mass of water, we use the equation:

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Volume of water = 35 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{35mL}\\\\\text{Mass of water}=(1g/mL\times 35mL)=35g$

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of metal = 4.82 g

$m_2$ = mass of water = 35 g

$T_{final}$ = final temperature = 34.5°C

$T_1$ = initial temperature of metal = 115°C

$T_2$ = initial temperature of water = 28.7°C

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$4.82\times c_1\times (34.5-110)=-[35\times 4.186\times (34.5-28.7)]$

$c_1=2.34J/g^oC$

Hence, the specific heat of metal is 2.34 J/g°C

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2 years ago

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Answer:

The concentration after 20 mins is 0.832 M

Explanation:

Zero order rate law is given by;

R = K [A₀]⁰

A zero order reaction, rate is independent of the initial concentration

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Where;

R is the rate of reaction

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Since R = K,

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After 20 min, the concentration will be;

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