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velikii [3]
3 years ago
12

Which linkages in starch and triglycerides result from condensation reactions? glycosidic bonds in starch and ester bonds in tri

glycerides ester bonds in both starch and triglycerides monosaccharide bonds in starch and glycerol bonds in triglycerides polar covalent bonds in starch and hydrophobic bonds in triglycerides?
Chemistry
1 answer:
bogdanovich [222]3 years ago
7 0
Glycosidic bonds in starch and ester bonds in triglycerides. The glycosidic bond is considered to be the covalent synthetic bonds that connection ring-molded sugar particles to different atoms. The frame by a buildup response between a liquor or amine of one particle and the anomeric carbon of the sugar, and hence, might be O-connected or N-connected.
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A stone is dropped from the roof of a building; 2.00s after that, a second stone is thrown straight down with an initial speed o
Brut [27]
A) -0.5(9.8)*t^2 = -25(t-2) - 0.5(9.8)(t-2)^2 
-4.9t^2 = -25t + 50 - 4.9(t^2-4t+4) 
0 = -25t+50+19.6t - 19.6 
5.4t = 30.4 
t = 5.62962963 s 

b) h = -4.9(5.62962963)^2 
h = -155.2943759 
the building is 155.2943759 m high 

c) speed 0of first stone 
= at 
= 9.8*5.62962963 
= 55.17037037 m/s 
speed of second stone
= v + at
= 25+9.8*3.62962963 
= 60.57037037 m/s
7 0
3 years ago
Which of the following combustion reactions is balanced correctly? A. C4H6 + 5.5O2 4CO2 + 3H2O B. C4H6 + 4O2 4CO2 + 3H2O C. C4H6
Airida [17]
You must verify that the number of atoms of each type is equal on both sides of the chemical equation: same number of C, same number of H and same number of O on both sides.

<span>A. C4H6 + 5.5O2 ---> 4CO2 + 3H2O

element      reactant side      product side

C                4                        4
H                6                        3*2 = 6
O                5.5 * 2 = 11        4*2 + 3 = 11

Then, this equation is balanced.

</span>Do the same with the other equations if you want to verify that they are not balanced.

Answer: option A.
5 0
3 years ago
The volume of a given mass of agas is 360cm cubic at 50 degrees and 700 milimetres Hg .find it's volume at s.t.p​
saveliy_v [14]

Answer:

V₂  =279.9 cm³

Explanation:

Given data:

Initial volume = 360 cm³

Initial temperature = 50°C

Initial pressure = 700 mmHg

Final volume = ?

Final temperature = 273 k

Final pressure = 1 atm

Solution:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Solution:

<em>We will convert the mmHg to atm.</em>

700/760 = 0.92 atm

<em>and °C to kelvin.</em>

50+273 = 323 K

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁ T₂/ T₁ P₂

V₂ = 0.92 atm × 360 cm³ × 273 K / 323 K ×1 atm

V₂  =  290417.6 atm .cm³.  K  / 323 k. atm

V₂  =279.9 cm³

8 0
3 years ago
Henry Moseley organized the periodic table by A. atomic weight. B. atomic number. C. electronegativity. D. overall charge of ato
Leni [432]
B - Atomic number. Dmitri Mendeleev organised the table according to atomic weight, however this caused problems with elements such as iodine and tellurium, Iodine has a higher mass, but a lower atomic number. And to make iodine in the same group as similar elements (halogens), Mendeleev had to break his own rules and put it before tellurium in the table. Moseley fixed this problem by ordering the elements according to atomic (proton) number.
3 0
3 years ago
Read 2 more answers
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
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