Answer:
1.The electonic configuration of elements and their position in the periodic table are related to each other, From the electronic configuration of the elements, we can determine the period and the group to which the element belongs
Let's consider, sodium with atomic number 11 and k, l, and M shells have 2,8,and 1 electrons. since, there are 3 principal energy levels so we concluded sodium belongs to third period M Shell(valance shell) has only 1 electrons. so sodium belongs to group 1.
2. Entire D-block elements are known as Transition Elements.
3. Group 17 is the halogen group.
4. Main group of elements are...... 1,2, and 13 through 18.
5. Group 18 are the noble gas elements .
12. a). Smaller
b). Increases
c). More reactive
d). Softer
7. a). k › Ca › Ge › Br › Kr
b). Ra › Ba › Sr › Ca › Mg › Be
9. a). Ca(calcium) ion is smaller.
b). Cl(chlorine) atom is smaller.
c). Mg(magnesium) atom is smaller.
10. a). F(fluorine)
b). Sr(strontium)
c). Pb(lead)
d). At(Astatine)
Answer: 143.3
137+115+104+263+98 divided by 5.Which gives us 143.3
The mass of Ba(IO3)2 that can be dissolved in 500 ml of water at 25 degrees celcius is 2.82 g
<h3>What mass of Ba(IO3)2 can be dissolved in 500 ml of water at 25 degrees celcius?</h3>
The Ksp of Ba(IO3)2 = 1.57 × 10^-9
Molar mass of Ba(IO3)2 = 487 g/mol?
Dissociation of Ba(IO3)2 produces 3 moles of ions as follows:
![Ksp = [Ba^{2+}]*[IO_{3}^{-}]^{2}](https://tex.z-dn.net/?f=Ksp%20%3D%20%5BBa%5E%7B2%2B%7D%5D%2A%5BIO_%7B3%7D%5E%7B-%7D%5D%5E%7B2%7D)
![[Ba(IO_{3})_{2}] = \sqrt[3]{ksp} =\sqrt[3]{1.57 \times {10}^{ - 9} } \\ [Ba(IO_{3})_{2}] = 1.16 \times {10}^{-3} moldm^{-3}](https://tex.z-dn.net/?f=%5BBa%28IO_%7B3%7D%29_%7B2%7D%5D%20%3D%20%20%5Csqrt%5B3%5D%7Bksp%7D%20%3D%3C%2Fp%3E%3Cp%3E%5Csqrt%5B3%5D%7B1.57%20%5Ctimes%20%20%7B10%7D%5E%7B%20-%209%7D%20%7D%20%5C%5C%20%20%5BBa%28IO_%7B3%7D%29_%7B2%7D%5D%20%3D%201.16%20%5Ctimes%20%20%7B10%7D%5E%7B-3%7D%20moldm%5E%7B-3%7D)
moles of Ba(IO3)2 = 1.16 × 10^-3 × 0.5 = 0.58 × 10^-3 moles
mass of Ba(IO3)2 = 0.58 × 10^-3 moles × 487 = 2.82 g
Therefore, mass Ba(IO3)2 that can be dissolved in 500 ml of water at 25 degrees celcius is 2.82 g.
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<h3>
Answer:</h3>
0.424 J/g °C
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality<u>
</u>
<u>Chemistry</u>
<u>Thermochemistry</u>
Specific Heat Formula: q = mcΔT
- q is heat (in Joules)
- m is mass (in grams)
- c is specific heat (in J/g °C)
- ΔT is change in temperature
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] m = 38.8 g
[Given] q = 181 J
[Given] ΔT = 36.0 °C - 25.0 °C = 11.0 °C
[Solve] c
<u>Step 2: Solve for Specific Heat</u>
- Substitute in variables [Specific Heat Formula]: 181 J = (38.8 g)c(11.0 °C)
- Multiply: 181 J = (426.8 g °C)c
- [Division Property of Equality] Isolate <em>c</em>: 0.424086 J/g °C = c
- Rewrite: c = 0.424086 J/g °C
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.424086 J/g °C ≈ 0.424 J/g °C
Answer:
The answer to your question is
1.- Volume = 3.4 ml
2.- Volume = 0.61 ml
3.- Mass = 2872.8 pounds
Explanation:
Problem 1
Volume = 18 ml
mass = 35.6 g
density = 10.5 g/ml
Process
1.- Calculate the volume of silver
Formula
solve for volume
Substitution
<u>volume = 3.4 ml</u>
2.- Problem 2
Total volume = ?
Volume = 18 + 3.4
Volume = 21.4 ml
Data
mass = 8.3 g
density = 13.6 g(ml
volume = ?
Formula
Solve for volume
Substitution
Result
<u>volume = 0.61 ml</u>
3.- Problem 3
Data
volume = 345 gal
density = 1 g/ml
mass = ?
Formula
Solve for mass
mass = density x volume
Covert gal to ml
1 gal --------------- 3785 ml
345 gal ------------- x
x = (345 x 3785) / 1
x = 1305825 ml
Substitution
mass = 1 x 1305825
mass = 1305825 g
Convert g to pounds
1 g ------------------- 0.0022 pounds
1305825 g ---------------- x
x = (1305825 x 0.0022)
<u> x = 2872.8 pounds</u>
