Heat water; mechanical (the movement of a turbine is based off of mechanical energy, not chemical or potential).
The temperature is 370K.
The volume of a given fuel pattern is immediately proportional to its absolute temperature at regular pressure (Charles's law). The volume of a given amount of fuel is inversely proportional to its pressure whilst temperature is held steady (Boyle's regulation).
Density is immediately proportional to stress and indirectly proportional to temperature. As stress increases, with temperature constant, density will increase. Conversely when temperature increases, with strain regular, density decreases.
The equations describing those legal guidelines are unique cases of the best gasoline regulation, PV = NRT, wherein P is the pressure of the gas, V is its extent, n is the number of moles of the gas, T is its kelvin temperature, and R is the ideal (common) gas constant.
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<u>Answer:</u> The rate law of the reaction is ![\text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BHgCl_2%5D%5BC_2O_4%5E%7B2-%7D%5D%5E2)
<u>Explanation:</u>
Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
For the given chemical equation:

Rate law expression for the reaction:
![\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BHgCl_2%5D%5Ea%5BC_2O_4%5E%7B2-%7D%5D%5Eb)
where,
a = order with respect to 
b = order with respect to 
Expression for rate law for first observation:
....(1)
Expression for rate law for second observation:
....(2)
Expression for rate law for third observation:
....(3)
Expression for rate law for fourth observation:
....(4)
Dividing 2 from 1, we get:

Dividing 2 from 3, we get:

Thus, the rate law becomes:
![\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BHgCl_2%5D%5E1%5BC_2O_4%5E%7B2-%7D%5D%5E2)
Answer: 2.8 moles
Explanation:
The balanced equation below shows that 1 mole of sodium oxide reacts with 1 mole of water to form 2 moles of sodium hydroxide respectively.
Na2O + H2O --> 2NaOH
1 mole of H2O = 2 moles of NaOH
Let Z moles of H2O = 5.6 mole of NaOH
To get the value of Z, cross multiply
5.6 moles x 1 mole= Z x 2 moles
5.6 = 2Z
Divide both sides by 2
5.6/2 = 2Z/2
2.8 = Z
Thus, 2.8moles of H2O are needed to produce 5.6 mol of NaOH
The new volume when pressure increases to 2,030 kPa is 0.8L
BOYLE'S LAW:
The new volume of a gas can be calculated using Boyle's law equation:
P1V1 = P2V2
Where;
- P1 = initial pressure (kPa)
- P2 = final pressure (kPa)
- V1 = initial volume (L)
- V2 = final volume (L)
According to this question, a 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, the volume is calculated as:
406 × 4 = 2030 × V2
1624 = 2030V2
V2 = 1624 ÷ 2030
V2 = 0.8L
Therefore, the new volume when pressure increases to 2,030 kPa is 0.8L.
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