If 3.00 g of limestone reacted, what mass of calcium chloride would be produced?

Write the balanced chemical equations for the decomposition of solid calcium hydroxide into solid calcium (ii) oxide (lime) and

Lesechka [4]

Calcium oxide (CaO) or lime in solid form can be prepared from the decomposition of calcium hydroxide {Ca $(OH)_{2}$ in to lime (CaO) and water ( $H_{2}O$ at high temperature. The reaction is an endothermic reaction. That is heat is absorbed in this reaction process. One mole of calcium hydroxide decomposed into one mole of calcium oxide and one mole of water. The balanced reaction can be shown as-CaC $O_{3}$ (solid) → CaO (solid) + $H_{2}O$ (liquid). The heat of the reaction is (+) 63.7 kJ/mole of CaO.

3 0

3 years ago

Explain how swirling occurs in mixing

TEA [102]

Explanation:

With respect to the mass transport mechanism in mixing, two fluids flow tangentially into the chamber and the radial component of velocity grows because fluids tend to flow through the outlet channel. These two components result in an overall rotating flow field inside the chamber

5 0

3 years ago

How many kilojoules of energy would be required to heat a 225g block of aluminum from 23.0 C to 73.5 C?

gulaghasi [49]

Answer:

$\boxed {\boxed {\sf 10.2 \ kJ}}$

Explanation:

We are asked to find how many kilojoules of energy would be required to heat a block of aluminum.

We will use the following formula to calculate heat energy.

$q=mc \Delta T$

The mass (m) of the aluminum block is 225 grams and the specific heat (c) is 0.897 Joules per gram degree Celsius. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature.

ΔT = final temperature - inital temperature

The aluminum block was heated from 23.0 °C to 73.5 °C.

ΔT= 73.5 °C - 23.0 °C = 50.5 °C

Now we know all three variables and can substitute them into the formula.

m= 225 g
c= 0.897 J/g° C
ΔT= 50.5 °C

$q= (225 \ g )(0.897 \ J/g \textdegree C)(50.5 \textdegree C)$

Multiply the first two numbers. The units of grams cancel.

$q= (225 \ g * 0.897 \ J/g \textdegree C)(50.5 \textdegree C)$

$q= (225 * 0.897 \ J / \textdegree C)(50.5 \textdegree C)$

$q= (201.825\ J / \textdegree C)(50.5 \textdegree C)$

Multiply again. This time, the units of degrees Celsius cancel.

$q= 201.825 \ J * 50.5$

$q= 10192.1625 \ J$

The answer asks for the energy in kilojoules, so we must convert our answer. Remember that 1 kilojoule contains 1000 joules.

$\frac { 1 \ kJ}{ 1000 \ J}$

Multiply by the answer we found in Joules.

$10192.1625 \ J * \frac{ 1 \ kJ}{ 1000 \ J}$

$10192.1625 * \frac{ 1 \ kJ}{ 1000 }$

$\frac {10192. 1625}{1000} \ kJ$

$10.1921625 \ kJ$

The original values of mass, temperature, and specific heat all have 3 significant figures, so our answer must have the same. For the number we found, that is the tneths place. The 9 in the hundredth place tells us to round the 1 up to a 2.

$10.2 \ kJ$

Approximately <u>10.2 kilojoules</u> of energy would be required.

3 0

3 years ago

Is anyone willing to help me with my chemistry?):

Neporo4naja [7]

I’m taking chem rn in 9th grade so I might be able to help. What’s the question?

3 0

3 years ago

What is the stationary phase in chromatography?

trapecia [35]

These are the two phases involved in chromatography. The mobile phase is usually liquid and is the solvent between the two. The solvent can attract some substances that has an affinity to it, thereby separating the polar from the non polar substances. On the other hand, the stationary phase is the media or medium which is a solid substance in which the unknown sample or the solution to be analysed is passed. As the term implies , it is stationary and substances in the unknown sample that has affinity to it or to the reagents in it will not travel along the column, but stay with the stationary medium, thereby allowing a separation of the substances present in the unknown sample/solution.

4 0

3 years ago