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Musya8 [376]
3 years ago
8

Question 4

Chemistry
1 answer:
padilas [110]3 years ago
5 0

Answer:

high wind speeds caused by seasonal storms

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Anyone knows gunpowder formula?​
ira [324]
Yeah , balanced, but still simplified, equation is: 10 KNO3 + 3 S + 8 C → 2 K2CO3 + 3 K2SO4 + 6 CO2 + 5 N2
5 0
3 years ago
Read 2 more answers
Calculate the change in entropy when 10.0 g of CO2 isothermally expands from a volume of 6.15 L to 11.5 L. Assume that the gas b
USPshnik [31]

Answer:

The change in entropy of the carbon dioxide is 1.183\times 10^{-3} kilojoules per Kelvin.

Explanation:

By assuming that carbon dioxide behaves ideally, the change in entropy (\Delta S), measured in kilojoules per Kelvin, is defined by the following expression:

\Delta S = m\cdot \bar c_{v}\cdot \ln \frac{T_{f}}{T_{o}}+m\cdot \frac{R_{u}}{M}\cdot \ln \frac{V_{f}}{V_{o}} (1)

Where:

m - Mass of the gas, measured in kilograms.

\bar c_{v} - Isochoric specific heat of the gas, measured in kilojoules per kilogram-Kelvin.

T_{o}, T_{f} - Initial and final temperatures of the gas, measured in Kelvin.

V_{o}, V_{f} - Initial and final volumes of the gas, measured in liters.

R_{u} - Ideal gas constant, measured in kilopascal-cubic meter per kilomole-Kelvin.

M - Molar mass, measured in kilograms per kilomole.

If we know that T_{o} = T_{f}, m = 0.010\,kg, R_{u} = 8.315\,\frac{kPa\cdot m^{3}}{kmol\cdot K}, M = 44.010\,\frac{kg}{kmol}, V_{o} = 6.15\,L and V_{f} = 11.5\,L, then the change in entropy of the carbon dioxide is:

\Delta S = \left[\frac{ (0.010\,kg)\cdot \left(8.315\,\frac{kPa\cdot m^{3}}{kmol\cdot K} \right)}{44.010\,\frac{kg}{kmol} } \right]\cdot \ln \left(\frac{11.5\,L}{6.15\,L}\right)

\Delta S = 1.183\times 10^{-3}\,\frac{kJ}{K}

The change in entropy of the carbon dioxide is 1.183\times 10^{-3} kilojoules per Kelvin.

4 0
3 years ago
An aqueous solution of potassium carbonate combine with a solution of calcium nitrate. What are the total and net ionic equation
romanna [79]
<span><span>K_2</span>C<span>O_3</span>(aq)+Ca(N<span>O_3</span><span>)_2</span>(aq)→ ?</span>
If we break these two reactants up into their respective ions, we get...<span><span>
K^+ </span>+ C<span>O^2_3 </span>+ C<span>a^<span>2+ </span></span>+ N<span>O_−3</span></span>
If we combine the anion of one reactant with the cation of the other and vice-versa, we get...<span>
CaC<span>O_3 </span>+ KN<span>O_3</span></span>

Now we need to ask ourselves if either of these is soluble in water. Based on solubility rules, we know that all nitrates are soluble, so the potassium nitrate is. Alternatively, we know that all carbonates are insoluble except those of sodium, potassium, and ammonium; therefore, this calcium carbonate is insoluble. This is good. It means we have a driving force for the reaction! That driving force is that a precipitate will form. In such a case, a precipitation reaction will occur, and the total equation will be...<span><span>
K_2</span>C<span>O_3</span>(aq) + Ca(N<span>O_3</span><span>)_2</span>(aq) → CaC<span>O_3</span>(s) + 2KN<span>O_3</span>(aq)</span>

To determine the net ionic equation, we need to remove all ions that appear on both sides of the equation in aqueous solution -- these ions are called spectator ions, and do not actually undergo any chemical reaction. To determine the net ionic equation, let's first rewrite the equation in terms of ions...
2K^+(aq) + CO_3^{2-}(aq) + Ca^{2+}(aq) + 2NO_3^{-}(aq) → Ca^{2+}(s) + CO_3^{2-}(s) + 2K^+(aq) + 2NO_3^-(aq)

The species that appear in aqueous solution on both sides of the equation (spectator ions) are... <span>
2K^+,NO_3^-</span>
If we remove these spectator ions from the total equation, we will get the net ionic equation...
CO_3^{2-}(aq) + Ca^{2+}(aq) <span>→</span> CaCO_3(s)
5 0
3 years ago
Increasing the temperature increases the rate of a many reactions by:
Bogdan [553]
Increasing the temperature increases reaction rates because of the disproportionately large increase in the number of high energy collisions. It is only these collisions (possessing at least the activation energy for the reaction) which result in a reaction.
3 0
3 years ago
What volume would 3.01•1023 molecules of oxygen gas occupy at STP?
frosja888 [35]
<h3>Answer:</h3>

                  Volume  = 11.2 L

<h3>Explanation:</h3>

Step 1: Calculate Moles:

                      As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.

The relation between Moles, Number of Particles and Avogadro's Number is given as,

                          Number of Moles  =  Number of Particles ÷ 6.022 × 10²³

Putting values,

                          Number of Moles  =  3.01× 10²³ Particles ÷ 6.022 × 10²³

                          Number of Moles  =  0.50 Moles

Step 2: Calculate Volume:

As we know that one mole of any Ideal gas at standard temperature and pressure occupies exactly 22.4 dm³ volume.

When 1 mole gas occupies 22.4 dm³ at STP then the volume occupied by 0.50 moles of gas is calculated as,

                      = (22.4 dm³ × 0.50 moles) ÷ 1 mole

                      = 11.2 dm³                                       ∴ 1dm³ = 1 L

So,

                                      Volume  = 11.2 L

4 0
4 years ago
Read 2 more answers
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