Answer:
300000Pa or 3×10^5 Pa
Explanation:
Since the problem involves only two parameters of volume and pressure, the formula for Boyle's law is suitably used.
Using Boyle's law
P1V1 = P2V2
P1 is the initial pressure = 1.5×10^5Pa
V1 is the initial volume = 0.08m3
P2 is the final pressure (required)
V2 is the final volume = 0.04 m3
From the formula, P2 = P1V1/V2
P2 = 1.5×10^5 × 0.08 ÷ 0.04
= 300000Pa or 3×10^5 Pa.
Omg i lost everything ugh
To do it again
1. 12g+2(16g)= 44g/mol
25.01/ 44g/mol= .... mol
2. 14g+3(1g)= 17g/mol
34.05g/ 17g/mol=.... mol
3. 23g+1g+ 12g+ 3(16g)= 84g/mol
17.31g/ 84g/mol=.... mol
4. 6(12g)+12(1g)+6(16g)= 180g/mol
123.44g/ 180g/mol=.... mol
5. 23g+16g+1g= 40g/mol
2.2mol x 40g/mol= .... g
6. 2(35g)= 71g/mol
4.5mol x 71g/mol= .... g
7. 137g+ 2(14g)+ 6(16g)= 261g/mol
0.002mol x 261g/mol= ....g
8. 2(56g)+ 3(32g)+ 12(16g)= 400g/mol
5.4mol x 400g/mol=.... g
I cant believe i had to do this all over
Answer:
C
Explanation:
a variable does that dose not depend on that of another
Answer: The initial temperature of the iron was 
Explanation:
As we know that,
![m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
.................(1)
where,
q = heat absorbed or released
= mass of iron = 360 g
= mass of water = 750 g
= final temperature = 
= temperature of iron = ?
= temperature of water = 
= specific heat of iron = 
= specific heat of water= 
Now put all the given values in equation (1), we get
![-360\times 0.450\times (46.7-x)=[750\times 4.184\times (46.7-22.5)]](https://tex.z-dn.net/?f=-360%5Ctimes%200.450%5Ctimes%20%2846.7-x%29%3D%5B750%5Ctimes%204.184%5Ctimes%20%2846.7-22.5%29%5D)
Therefore, the initial temperature of the iron was
Answer:
250,000 ML is equal to 250 Liters.
Explanation:
The term 'Mili' multiplies the term 'Liter' by 1,000. Per every liter is 1000 milliliters.
