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Usimov [2.4K]
2 years ago
11

The density of whole blood is 1.05 g/ml. a typical adult has between 4.7 and 5.5 l of whole blood. what is the mass in pounds of

4.7 l of whole blood?
Chemistry
1 answer:
Kisachek [45]2 years ago
5 0
<span>1.05 g/ml * 1000 ml = 1050g/l because of 1g/ml = 1 kg/l so, a/q mass of 4.7 l of whole blood in pound = 4.7 * 1050 = 4935 g so in pound 4935g = 10.87981p</span>
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pogonyaev

Answer:

1)The proximity of the positively charged phosphorous and negatively charged carbon stabilizes the charges.

2) Inductive effects and resonance stabilize the negative charge

Explanation:

both atoms have full octets of electrons( I.e Carbon and say phosphorus). The result can be viewed as a structure in which two adjacent atoms are connected by both a covalent and an ionic bond; normally written X+–Y−. Ylides are thus 1,2-dipolar compounds, and a subclass of zwitterions

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2 years ago
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You can purchase nitric acid in a concentrated form that is 70.3% HNO3 by mass and has a density of 1.41 g/mL. Describe exactly
Andrej [43]
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Standard Thermodynamic Quantities for Selected Substances at 25 ∘C Reactant or product ΔHf∘(kJ/mol) Al(s) 0.0 MnO2(s) −520.0 Al2
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Answer:

-1815.4 kJ/mol

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This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.

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4 0
2 years ago
If the value of Kc for the reaction is 2.50, what are the equilibrium concentrations if the reaction mixture was initially 0.500
ratelena [41]

<u>Answer:</u> The equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.

<u>Explanation:</u>

We are given:

Initial concentration of sulfur dioxide = 0.500 M

Initial concentration of nitrogen dioxide = 0.500 M

Initial concentration of sulfur trioxide = 0.00500 M

Initial concentration of nitrogen monoxide = 0.00500 M

The chemical reaction follows:

                         SO_2+NO_2\rightleftharpoons SO_3+NO

<u>Initial:</u>             0.500  0.500      0.005   0.005

<u>At eqllm:</u>      0.500-x  0.500-x   0.005+x  0.005+x

The expression of equilibrium constant for the above reaction follows:

K_c=\frac{[SO_3][NO]}{[SO_2][NO_2]}

We are given:

K_c=2.50

Putting values in above equation, we get:

2.50=\frac{(0.005+x)\times (0.005+x)}{(0.500-x)\times (0.500-x)}\\\\x=0.304,1.37

Neglecting the value of x = 1.37, because change cannot be greater than the initial concentration

So, equilibrium concentration of sulfur dioxide = (0.500-x)=(0.500-0.304)=0.196M

Equilibrium concentration of nitrogen dioxide = (0.500-x)=(0.500-0.304)=0.196M

Equilibrium concentration of sulfur trioxide = (0.00500+x)=(0.00500+0.304)=0.309M

Equilibrium concentration of nitrogen monoxide = (0.00500+x)=(0.00500+0.304)=0.309M

Hence, the equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.

4 0
2 years ago
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