Answer:
900 K
Explanation:
Recall the ideal gas law:

Because only pressure and temperature is changing, we can rearrange the equation as follows:

The right-hand side stays constant. Therefore:

The can explodes at a pressure of 90 atm. The current temperature and pressure is 300 K and 30 atm, respectively.
Substitute and solve for <em>T</em>₂:

Hence, the temperature must be reach 900 K.
Answer:
The correct answer is pOH= 11
Explanation:
From the aqueous acid-base equilibrium we know that
pH + pOH = 14
If we know pH, we can calculate pOH as follows:
pOH = 14 - pH
In this problem, the solution has a pH of 3, so:
pOH = 14 - 3 = 11
They travel through all medium types
Answer:- 13.6 L
Solution:- Volume of hydrogen gas at 58.7 Kpa is given as 23.5 L. It asks to calculate the volume of hydrogen gas at STP that is standard temperature and pressure. Since the problem does not talk about the original temperature so we would assume the constant temperature. So, it is Boyle's law.
Standard pressure is 1 atm that is 101.325 Kpa.
Boyle's law equation is:

From given information:-
= 58.7 Kpa
= 23.5 L
= 101.325 Kpa
= ?
Let's plug in the values and solve it for final volume.

On rearranging the equation for 

= 13.6 L
So, the volume of hydrogen gas at STP for the given information is 13.6 L.
A. NaCl(s) and O2(g)
B. 2NaClO3(s) —> 2NaCl(s) + 3O2(g)
C. moles NaClO3 = 100 g / 106.44 g/mol = 0.939 mol NaClO3
D. 0.939 mol NaCl (because the NaClO3 and NaCl are in a 1 to 1 ratio)
E. grams NaCl = 0.939 mol • 58.44 g/mol = 54.9 g NaCl
F. moles of O2 = 0.939 mol NaClO3 • (3 mol O2 / 2 mol NaClO3) = 1.41 mol O2
G. grams of O2 = 1.41 mol • 32 g/mol = 45.1 g O2
H. Percent yield = 10/45.1 • 100% = 22.2% yield