Answer:
Approximately 
, if this gas is an ideal gas, and that the quantity of this gas stayed constant during these changes.
Explanation:
Let 
and 
denote the pressure of this gas before and after the changes.
Let 
and 
denote the volume of this gas before and after the changes.
Let 
and 
denote the temperature (in degrees Kelvins) of this gas before and after the changes.
Let 
and 
denote the quantity (number of moles of gas particles) in this gas before and after the changes.
Assume that this gas is an ideal gas. By the ideal gas law, the ratios 
and 
should both be equal to the ideal gas constant, 
.
In other words:
.
.
Combine the two equations (equate the right-hand side) to obtain:
.
Rearrange this equation for an expression for , the temperature of this gas after the changes:
.
Assume that the container of this gas was sealed, such that the quantity of this gas stayed the same during these changes. Hence: 
, 
.
![\begin{aligned} T_2 &= \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2}\cdot T_1 \\[0.5em] &= \frac{1.67\; \rm atm}{1.12\; \rm atm} \times \frac{11.2\; \rm m^{3}}{15.7\; \rm m^{3}} \times 1 \times 245\; \rm K \\[0.5em] &\approx 261\; \rm K\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20T_2%20%26%3D%20%5Cfrac%7BP_2%7D%7BP_1%7D%20%5Ccdot%20%5Cfrac%7BV_2%7D%7BV_1%7D%20%5Ccdot%20%5Cfrac%7Bn_1%7D%7Bn_2%7D%5Ccdot%20T_1%20%5C%5C%5B0.5em%5D%20%26%3D%20%5Cfrac%7B1.67%5C%3B%20%5Crm%20atm%7D%7B1.12%5C%3B%20%5Crm%20atm%7D%20%5Ctimes%20%5Cfrac%7B11.2%5C%3B%20%5Crm%20m%5E%7B3%7D%7D%7B15.7%5C%3B%20%5Crm%20m%5E%7B3%7D%7D%20%5Ctimes%201%20%5Ctimes%20245%5C%3B%20%5Crm%20K%20%5C%5C%5B0.5em%5D%20%26%5Capprox%20261%5C%3B%20%5Crm%20K%5Cend%7Baligned%7D)
.
Answer:
Only the goalie is allowed inside the goal crease. The only exception when another player is allowed in the goal area is when they take off from outside the goal area, and shoots or passes the ball before landing. To avoid interference with other players, the player must then exit the goal area as soon as possible.
Explanation:
Answer:
The answer to your question is KNO₃
Explanation:
Data
HNO₃
KOH
neutralization reaction
Process
1.- In a neutralization reaction an acid reacts with an alkali and the products are water and a binary or ternary salt.
Reactants acid = HNO₃ alkali = KOH
A neutralization reaction is a double displacement reaction so the salt formed is.
HNO₃ + KOH ⇒ H₂O + KNO₃
Potassium nitrate
Answer:
750cm³
Explanation:
volume = length × width × height
15 × 10 × 5 = 750
750cm³
