The solute has to be hydrophilic, (water loving).
Answer:
467
Explanation:
ncl2 = 454.4x1/(71.0 g/mol) = 6.40 mols cl2
6.40 mols cl2 x 2molsHCL/1moleCL2 x 36.5g/1moleHCL = <u>467 g HCL</u>
Answer:
2B2 + 3O2 → 2B2O3
Explanation:
Balance The Equation: B2 + O2 = B2O3
1. Label Each Compound With a Variable
aB2 + bO2 = cB2O3
2. Create a System of Equations, One Per Element
B: 2a + 0b = 2c
O: 0a + 2b = 3c
3. Solve For All Variables (using substitution, gauss elimination, or a calculator)
a = 2
b = 3
c = 2
4. Substitute Coefficients and Verify Result
2B2 + 3O2 = 2B2O3
L R
B: 4 4 ✔️
O: 6 6 ✔️
hope this helps!
Answer:
1.07 g
Explanation:
Half-life of Pu-234 = 4.98 hours
Initially present = 45 g
mass remains after 27 hours = ?
Solution:
Formula
mass remains = 1/ 2ⁿ (original mass) ……… (1)
Where “n” is the number of half lives
To find "n" for 27 hours
n = time passed / half-life . . . . . . . .(2)
put values in equation 2
n = 27 hr / 4.98 hr
n = 5.4
Mass after 27 hr
Put values in equation 1
mass remains = 1/ 2ⁿ (original mass)
mass remains = 1/ 2^5.4 (45 g)
mass remains = 1/ 42.2 (45 g)
mass remains = 0.0237 x 45 g
mass remains = 1.07 g
