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Sav [38]
3 years ago
10

What is the reasoning behind Forgacs’s ideas onbiofabrication?

Chemistry
1 answer:
kondor19780726 [428]3 years ago
5 0
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When comparing the colors of the following four compounds, which is most likely to appear green? [Co(CN)6]3− [Co(H2O)6]3+ [Co(en
alexandr1967 [171]

Answer:

3

Explanation:

its bc

5 0
3 years ago
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Calculate the amount of heat needed to boil 159.g of water
MissTica
Okay I did the math and I'm guessing around 18*C
5 0
3 years ago
What is the right answer for this. I really need help
statuscvo [17]
<span>así que te está diciendo que</span>
6 0
3 years ago
How many CaF are in a 1.7x10^25 please I need help fast!!
Gemiola [76]
<h3>Answer:</h3>

28 mol CaF

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 1.7 × 10²⁵ molecules CaF

[Solve] moles CaF

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                          \displaystyle 1.7 \cdot 10^{25} \ molecules \ CaF(\frac{1 \ mol \ CaF}{6.022 \cdot 10^{23} \ molecules \ CaF})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 28.2298 \ moles \ CaF

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

28.2298 mol CaF ≈ 28 mol CaF

7 0
3 years ago
A gas mixture with 4 mol of Ar, x moles of Ne, and y moles
maks197457 [2]

Answer:

a) \Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

b) x=4

c) \Delta G_{max}=-2721.9 J/mol

Explanation:

Gas mixture:

n_{Ar}= 4 mol

n_{Ne}= x mol

n_{Xe}= y mol

n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}

n_{Ne} + n_{Xe}=2*n_{Ar}

x + y=8 mol

y=8 mol- x

Mol fractions:

x_{Ar}=\frac{4 mol}{12 mol}=1/3

x_{Ne}=\frac{x mol}{12 mol}=x/12

x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12

Expression of \Delta G_{mixing}

\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)

\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]

\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

Expression of \Delta G_{max}

\frac{d \Delta G_{mixing}}{dx}=0

\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]

0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)

0=[ln (x/12)-ln ((8-x)/12)

ln (x/12)=ln ((8-x)/12)

x=(8-x)

x=4

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]

\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]

\Delta G_{max}=-2721.9 J/mol

4 0
3 years ago
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