Answer:
7.22L
Explanation:
At Stp T2 = 273K and P2 is 1 atm
So we have the following data
V1 =10L. P1=650mmHg(0.855atm)
V2 =V2. P2 = 1 atm
T1=50°c (323K) T2 =273K
Hence from the combined gas law
P1xV1 /T1 =P2xV2/T2
0.855x10/323 =1xV2 /273
8.55/323 =V2 /273
V2 = 8.55x273 /323
V2 = 7.226 L
Answer:
f. Sn^4+
c. second
e. Al^3+
d. third
Explanation:
This question comes from a quantitative analysis showing the flowchart of a common scheme for identifying cations.
Now, from the separation scheme, Let's assume that Sn⁴⁺ & Al³⁺ were given; Then, Yes, the separation will work.
However, there will be occurrence of precipitation after the 1st step1.
So, the <u>Sn⁴⁺</u> cation will precipitate after the <u>second </u>step. Then the <u>Al³⁺</u> cation will precipitate after the <u>third</u> step.
Answer:
Explanation: A compound with the empirical formula SO has a molecular weight of 96.13
Answer:
2.5x10^–3 mole.
Explanation:
Data obtained from the question include:
Volume of solution = 25mL
Molarity of HNO3 = 0.1M
Mole of HNO3 =..?
First, we'll begin by converting 25mL to L. This can be achieved by doing the following:
1000mL = 1L
Therefore, 25mL = 25/1000 = 0.025L
Now, we can obtain the number of mole of HNO3 present in the solution as follow:
Molarity = mole /Volume
Mole = Molarity x Volume
Mole = 0.1 x 0.025
Mole = 2.5x10^–3 mole.
Therefore, 2.5x10^–3 mole of HNO3 is present in the solution.
Answer:
ob
Explanation:
becoz when volume increases density decreases.
