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rjkz [21]
2 years ago
10

The equation below represents chemical reaction that occurs in living cells. C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + energy How many atoms

are represented in the reactants of this equation?
Chemistry
1 answer:
KengaRu [80]2 years ago
4 0

Answer:

36

Explanation:

There are 6 Carbon atoms, 12 Hydrogen atoms, and 18 Oxygen atoms.

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A red blood cell is placed into each of the following solutions. Indicate whether crenation, hemolysis, or neither will occur.
krok68 [10]

Answer:

Following are the responses to the given choices:

Explanation:

  • The RBC crenation is implied through NaCl by 2,67 percent(m/v) because that solution becomes hypertonic to RBC because of the water within the RBC that passes externally towards the outskirts. RBC thus shrinks.
  • 1.13% (m/v), because the low concentration or osmotic that all this solution shows is hypotonic regarding RBC because of the water which has reached the resulting swelling in RBC.
  • Distilled H2 implies hemolytic distillation.
  • Glucose is indicated by crenation at 8.69 percent (m/v).
  • 5.0% (m/v) glucose and 0.9% (m/v) (Crenation is indicated by NaCl.v)
3 0
2 years ago
Identify the correct net ionic equation for the reaction that occurs when solutions of Pb(NO3)2 and NH4Cl are mixed.
Pie

Answer:

Pb2+(aq) + 2Cl–(aq) ----> PbCl2(s)

Explanation:

The net ionic equation shows the main reaction that takes place in a system. Hence, a net ionic equation focusses only on those species that actually participate in the reaction.

For the reaction between Pb(NO3)2 and NH4Cl , the net ionic equation is;

Pb^+(aq) + 2Cl^-(aq) ---> PbCl2(s)

7 0
3 years ago
The freezing-point depression of a 0.100 m MgSO4 solution is 0.225°C. Determine the experimental van't Hoff factor of MgSO4 at t
Andrews [41]

<u>Answer:</u> The experimental van't Hoff factor is 1.21

<u>Explanation:</u>

The expression for the depression in freezing point is given as:

\Delta T_f=iK_f\times m

where,

i = van't Hoff factor = ?

\Delta T_f = depression in freezing point  = 0.225°C

K_f = Cryoscopic constant  = 1.86°C/m

m = molality of the solution = 0.100 m

Putting values in above equation, we get:

0.225^oC=i\times 1.86^oC/m\times 0.100m\\\\i=\frac{0.225}{1.86\times 0.100}=1.21

Hence, the experimental van't Hoff factor is 1.21

7 0
3 years ago
There are two isotopes of an unknown element, X-19 and X-21. The abundance of X-19 is 14.55%. A weighted average uses the percen
alekssr [168]

Answer:

2.765amu is the contribution of the X-19 isotope to the weighted average

Explanation:

The average molar mass is defined as the sum of the molar mass of each isotope times its abundance. For the unknown element X that has 2 isotopes the weighted average is defined as:

X = Mass X-19 * Abundance X-19 + MassX-21 * Abundance X-21

The contribution of the X-19 isotope is its mass (19.00 amu) times its abundance (14.55% = 0.1455). That is:

19.00amu * 0.1455 =

2.765amu is the contribution of the X-19 isotope to the weighted average

8 0
2 years ago
The thallium Subscript 81 Superscript 208 Baseline Tl nucleus is radioactive, with a half-life of 3.053 min. At a given instant,
Genrish500 [490]

Answer: (E) 300 bq

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Radioactive decay process is a type of process in which a less stable nuclei decomposes to a stable nuclei by releasing some radiations or particles like alpha, beta particles or gamma-radiations. The radioactive decay follows first order kinetics.

Half life is represented by t_{\frac{1}{2}

Half life of Thallium-208 = 3.053 min

Thus after 9 minutes , three half lives will be passed, after ist half life, the activity would be reduced to half of original i.e. \frac{2400}{2}=1200, after second  half life, the activity would be reduced to half of 1200 i.e. \frac{1200}{2}=600,  and after third half life, the activity would be reduced to half of 600 i.e. \frac{600}{2}=300,

Thus the activity 9 minutes later is 300 bq.

7 0
2 years ago
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