Answer:
The answer to your question is 25.2 g of acetic acid.
Explanation:
Data
[Acetic acid] = 0.839 M
Volume = 0.5 L
Molecular weight = 60.05 g/mol
Process
1.- Calculate the number of moles of acetic acid
Molarity = moles / volume
-Solve for moles
moles = Molarity x volume
-Substitution
moles = (0.839)(0.5)
-Result
moles = 0.4195
2.- Calculate the mass of acetic acid using proportions and cross multiplications
60.05 g ----------------------- 1 mol
x ----------------------- 0.4195 moles
x = (0.4195 x 60.05) / 1
x = 25.19 g
3.- Conclusion
25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M
Hello.
If you wanted to harvest cranberries you would need to use flotation.
Have a nice day.
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!
Which of these is an isoelectronic series? 1) na+, k+, rb+, cs+ 2) k+, ca2+, or, s2– 3) na+, mg2+, s2–, cl– 4) li, be, b, c 5) n
ss7ja [257]
An isoelectronic series is where all of the ions listed have the same number of electrons in their atoms. When an atom has net charge of zero or neutral, it has equal number of protons and electrons. Hence, it means that the atomic number = no. of protons = no. of electrons. If these atoms become ions, they gain a net charge of + or -. Positive ions are cations. This means that they readily GIVE UP electrons, whereas negative ions (anions) readily ACCEPT electrons. So, to know which of these are isoelectronic, let's establish first the number of electron in a neutral atom from the periodic table:
Na=11; K=19; Rb=37; Cs = 55; Ca=20; S=16; Mg=12; Li=3; Be=4; B=5; C=6
A. Na⁺: 11-1 = 10 electrons
K⁺: 19 - 1 = 18 electrons
Rb⁺: 37-1 = 36 electrons
B. K⁺: 19 - 1 = 18 electrons
Ca²⁺: 20 - 2 = 18 electrons
S²⁻: 16 +2 = 18 electrons
C. Na⁺: 11-1 = 10 electrons
Mg²⁺: 12 - 2 = 10 electrons
S²⁻: 16 +2 = 18 electrons
D. Li=3 electrons
Be=4 electrons
B=5 electrons
C=6 electrons
The answer is letter B.