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11 months ago

Complete combustion of 2.0 metric tons of coal to gaseous carbon dioxide releases 6.6X10¹⁰ J of heat. Convert this energy to

Chemistry

1 answer:

Jobisdone [24]11 months ago

4 0

6.6 × $10^{10}$ J = 6.2 × $10^{7}$ British thermal units

First of all, we should know that 1 British thermal unit = 1055 Joules.

From question, heat = 6.6 × $10^{10}$ J

1 BTU = 1055 J

6.6 × $10^{10}$ J = $\frac{1}{1055}$ × 6.6 × $10^{10}$ British thermal units

= 6.2 × $10^{7}$ British thermal units

1000 Joules = 1 kilojoule

A kilojoule is a unit of measure of energy, in the equal way that kilometers degree distance. meals energy used to be measured in energy and some international locations still use those gadgets.

The power we get from foods and drinks is measured in kilojoules. that is the metric time period for calorie. Kilojoules and energy constitute the equal aspect. One calorie is ready four kilojoules.

Calorie a unit for measuring warmth same to the quantity of warmth required to elevate the temperature of one gram of water one diploma Celsius.

Learn more about kilojoules here :- brainly.com/question/490326

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when hydrogen atom absorbs a photon, and an electron moves level 1 to level 2, what happens to the energy of the atom?

pychu [463]

Answer:

When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. One way of thinking about this higher energy state is to imagine that the electron is now moving faster, (it has just been "hit" by a rapidly moving photon).

Explanation: pls mark brainliest :))

7 0

2 years ago

Need help ASAP!<br><br> How many moles of sodium nitrate are in 0.25 L of 1.2 M NaNO3 solution?

Setler79 [48]

Answer:

$\boxed {\boxed {\sf 0.3 \ mol \ NaNO_3}}$

Explanation:

Molarity is a measure of concentration in moles per liter.

$molarity= \frac{moles \ of \ solute}{liters \ of \ solution}$

The molarity of the solution is 1.2 M NaNO₃ or 1.2 moles NaNO₃ per liter. There are 0.25 liters of the solution. The moles of solute are unknown, so we can use x.

molarity= 1.2 mol NaNO₃/L
liters of solution=0.25 L
moles of solute =x

$1.2 \ mol \ NaNO_3/L= \frac{x}{0.25 \ L}$

We are solving for x, so we must isolate the variable, x. It is being divided by 0.25 liters. The inverse of division is multiplication, so we multiply both sides by 0.25 L.

$0.25 \ L *1.2 \ mol \ NaNO_3/L=\frac{x}{0.25 \ L} *0.25 \ L$

$0.25 \ L *1.2 \ mol \ NaNO_3/L=x$

The units of liters cancel, so we are left with the units moles of sodium nitrate.

$0.25 *1.2 \ mol \ NaNO_3=x$

$0.3 \ mol \ NaNO_3=x$

There are 0.3 moles of sodium nitrate.

3 0

2 years ago

The octane number of gasoline is a measure of octane activity. The octane activity is determined by comparing the engine perform

Semmy [17]

Answer:

A. True

Explanation:

The octane number is determined by comparing the characteristics of gasoline to isooctane (2,2,4-trimethylpentane) and heptane. The correct option is option A.

Basically, the higher the octane number, the greater the resistance of the gasoline to knocking.

4 0

2 years ago

How many kilocalories are required to increase the temperature of 15.6 g of iron from 122 °c to 355 °c. the specific heat of iro

Dmitriy789 [7]

Heat require to boil 15.6 g iron from 122 C0to 355 C0 whereas,

$Q = m s dT$

Where, m is mass of iron

s is specific heat of iron

d T is change in temperature in celcius

$= 15.6 g * 0.45 J /g /C * (355 - 122) = 1.63 * 10^3 J$

1 cal = 4.2 J

Then,

$Q = (1.63 * 10^3) /4.2 = 0.389 K cal$

Thus 0.389 k cal of enrgy is required by a 15.6 g Fe to reach to 355 C^0

4 0

3 years ago

For doping silicon with boron, silicon specimen was kept in gaseous atmosphere containing B2O3 that maintained the B concentrati

Kay [80]

Solution :

From Fick's law:

$\frac{D_{AB}}{\Delta z } \times (C_{A1}-C_{A2})=N_a$

Mass balance: Exits = Accumulation

$-N_A A = \frac{dm}{dt}$

$-N_A A = \frac{dVp}{dt}$

$-N_A A = \frac{dV}{dt}p$

$-N_A A = \frac{dhA}{dt}$

$-N_A A = \frac{dh}{dt} \times Ap$

From the last step, area cancels out and thus leaves :

$-N_A = \frac{dh}{dt} \times p$

So now we can substitute the $N_A$ by the Fick's law

$\frac{D_{AB}}{\Delta z } \times (C_{A1}-C_{A2})=\frac{dh}{dt} p$

Substituting the values we get

$=\frac{-4 \times 10^{-13}}{0.0001} \times (3 \times 10^{26} - C_{A2}) = \frac{dh}{dt} \times 2.46$

$=-4 \times 10^{-9} \times( 3 \times 10^{26} - C_{A2}) = 0.0001 \times 2.46$

$= -7800 \times 4 \times 10^{-9} \times (3 \times 10^{26}-C_{A2})=0.000246$ $=-(3 \times 10^{26}-C_{A2}) = 7.8846 \times 100 \times \frac{1}{69.62} \times 6.022 \times 10^{23}$

$C_{A2} = 6.85 \times 10^{28} \ \text{ boron atoms} /m^3$

3 0

2 years ago