So to solve this you need to know Charles’s law which is: V1/T1=V2/T2. Where T1 and V1 is the initial volume and Temperature and V2 and T2 is the temperature and volume afterwards. So first plug in the numbers you are given. V1= 1.55L T1= 32C° V2= 755mL T2=?. Since your volumes are two different units you change 755mL to be in L so that would be 0.755 L. And since your temp isn’t in Kelvin you do 273+32= 305K°. You then would rearrange your equation to solve for T2 which is V2T1/V1. Then you plug in your numbers (0.755L)(305K)/1.55L. Then you solve and would be 148.5645161 —> 1.49 x 10^2 K
Answer:
NH4Cl, NaCl, Ba(OH)2, NaOH
Explanation:
NH4Cl is an acidic salt formed by the neutralization of a strong acid (HCl) with a weak base (NH3). Hence, it will habe a PH<7 (the lowest PH).
NaCl is a neutral salt,formed by neutralization of a strong acid (HCl) with a strong base (NaOH). Hence, it will have a PH of 7.
Ba(OH)2 is a weak base. Therefore, it will have a PH between 8 and 10.
NaOh meanwhile, is a strong base. Therefore, it will have a PH between 10 to 13.
Hence, we have
NH4Cl < NaCl < Ba(OH)2 < NaOH
Answer:
The percent by mass of copper in the mixture was 32%
Explanation:
The ammount of HNO₃ used is:
mol HNO₃ = volume * concentration
mol HNO₃ = 0.015 l * 15.8 mol/l
mol HNO₃ = 0.237 mol
According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.
Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.
1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:
0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.
Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:
100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g
Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%
2HNO3 + Ca(OH)2 --> 2HOH + Ca(NO3)2
Answer:
Pressure gas A
using boyles law
= 
V2
V2 = 717ml + 179 ml
= 896ml
∴ = 2.50 × 717ml/896ml
= 2.0 bar
Pressure B
P2 = 4.30 bar× 179ml/896ml
= 0.859bar
ptotal = 
+
= 2.0 bar + 0.859 bar
= 2.859 bar
Explanation: Using Daltons law of partial pressure,the pressure is independently of each other when the gas is exerted.where we can use daltons law to find the pressure of each gas separately when it expands into the total volume in two containers.
