<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
Polar.
Polar bonds have unequal sharing electrons while nonpolar, the opposite, has equal sharing electrons. This is a tactic typically used to determine whether or not a compound or element itself is polar or nonpolar.
Hope this helps!
Answer:
First choice: 2
Explanation:
There are 2 phosphorous (P) in the substance.
Ignore the strontium (Sr3) part because you are looking to isolate the P from (PO4)2.
Break the chemical equation apart to get 1 Phosphorous atom, and 4 Oxygen atoms.
Now, multiple 1 by 2 because that are 2 phosphate to get 2 phosphorous atoms.
<h2>Answer with explanation </h2>
<h3><em>The starting diol for this molecule is :-</em></h3><h3><em>The starting diol for this molecule is :-D) ethan-1,2-diol.</em></h3>
<em>Hope </em><em>my </em><em>answer </em><em>is</em><em> helpful</em><em> to</em><em> you</em><em> </em><em>☺️</em>
Option (a) is correct.
A reducing agent is the one which loses electrons to other substance and an oxidizing agent is one which accepts electrons.
Here, In
, Cr has oxidation number 6+ in the L.H.S of the equation, but on R.H.S its oxidation number is 0 i.e. it Cr has gained electrons such that total charge is 0.
And the oxidation state of Al in the left-hand side of equation is 0 and in right-hand side, it is +6.i.e. it has donated its electrons to Cr.
Hence, Cr is the oxidizing agent and Al is the reducing agent.
