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RSB
1 month ago
7

in a typical person, the level of glucose is about 85 mg/ 100 mL of blood. if an average body contains about 11 pt of blood, how

many grams and how many pounds of glucose are present in the blood?
Chemistry
1 answer:
Anarel [2.9K]1 month ago
5 0

It's important to remember that 1 pint equals 473.1765 mL, therefore 11 pints amounts to 5204.9415 mL.

We can formulate a proportion based on the problem statement

(85 mg glucose/ 100 mL) times (1 g/ 1000 mg) = 4.4242 grams of glucose

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You have a racemic mixture of d-2-butanol and l-2-butanol. the d isomer rotates polarized light by +13.5∘. what is the rotation
Anarel [2989]
The L- isomer serves as the enantiomer of the D- isomer, and given that the optical rotation of the D- isomer is + 13.5°, the L- isomer's optical rotation will have the same magnitude but an opposite sign, resulting in -13.5°.

Thus, the rotation of the racemic mixture will be equal to 0°.


- This occurs because a racemic mixture contains equal proportions of both enantiomers.
8 0
2 months ago
Suppose you wanted to make a buffer of exactly ph 7.00 using kh2po4 and na2hpo4. if the final solution was 0.10 m in kh2po4, wha
Tems11 [2777]
The answer is - 0.138 M. The buffer pH can be determined using the Henderson equation. Here, KH_2PO_4 acts as a weak acid and Na_2HPO_4 serves as its corresponding conjugate base. The weak acid has two protons, while the base contains one. The equation can therefore be expressed in terms of protons transferred. Phosphoric acid can donate protons in three stages; the equation we’ve referenced pertains to the second stage, as the acid then has only two protons available and the base only one. Given the concentration of the acid as 0.10 M, we need to calculate the concentration of the base necessary to form a buffer with a pH of exactly 7.0. Substituting the values into the equation leads us to the solution. Cross-multiplying, we find that [base] = 1.38(0.10), yielding [base] = 0.138. Therefore, the concentration of the base needed for the buffer is 0.138 M.
5 0
1 month ago
A 20.0–milliliter sample of 0.200–molar K2CO3 so­lution is added to 30.0 milliliters of 0.400–mo­lar Ba(NO3)2 solution. Barium c
KiRa [2933]

Respuesta:

0.16 M

Explicación:

Teniendo en cuenta:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

O sea,

Moles =Molarity \times {Volume\ of\ the\ solution}

Dado que:

Para K_2CO_3 :

Molaridad = 0.200 M

Volumen = 20.0 mL

Convierte mL a L:

1 mL = 10⁻³ L

Entonces, volumen = 20.0×10⁻³ L

Los moles de K_2CO_3 son:

Moles=0.200 \times {20.0\times 10^{-3}}\ moles

Moles de K_2CO_3 = 0.004 moles

Para Ba(NO_3)_2 :

Molaridad = 0.400 M

Volumen = 30.0 mL

Convertimos mL a L:

1 mL = 10⁻³ L

Volumen = 30.0×10⁻³ L

Entonces, los moles de Ba(NO_3)_2 son:

Moles=0.400 \times {30.0\times 10^{-3}}\ moles

Moles de Ba(NO_3)_2 = 0.012 moles

Según la reacción:

Ba(NO_3)_2 + K_2CO_3\rightarrow BaCO_3 + 2KNO_3

1 mol de Ba(NO_3)_2 reacciona con 1 mol de K_2CO_3

Por lo tanto,

0.012 mol de Ba(NO_3)_2 reacciona con 0.012 mol de K_2CO_3

Moles disponibles de K_2CO_3 = 0.004 mol

El reactivo limitante es el que está en menor cantidad, entonces K_2CO_3 es el limitante (0.004 < 0.012).

La formación del producto depende del reactivo limitante, así que,

1 mol de K_2CO_3 reacciona con 1 mol de Ba(NO_3)_2 y produce 1 mol de BaCO_3

0.004 mol de K_2CO_3 reacciona con 0.004 mol de Ba(NO_3)_2 y genera 0.004 mol de BaCO_3

Los moles restantes de Ba(NO_3)_2 son: 0.012 - 0.004 = 0.008 mol

El volumen total es 20 + 30 mL = 50 mL = 0.050 L

Por lo que la concentración del ion bario, Ba^{2+}, después de la reacción es:

Molarity=\frac{0.008}{0.050}\ M = 0.16\ M

3 0
3 months ago
A chemist is looking for an element that reacts similarly to the element lithium (LI). Which would be the best choice?
lions [2927]
Option d is the correct choice, as both belong to the alkali metals category (group one).
8 0
2 months ago
Read 2 more answers
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