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Pani-rosa
1 month ago
15

The Thompson analogy titled "The Carpet-Seed Children Analogy" attempts to deal with the issue of failed ____________________(A)

fertilization(B) intercourse(C) conception(D) contraception
Chemistry
1 answer:
castortr0y [2.7K]1 month ago
8 0

Answer: (D) contraception

Explanation:

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doctor has ordered that a patient be given 20 g of glucose, which is available in a concentration of 70.00 g glucose/1000.0 mL o
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The patient needs to receive 285.71 ml.
1000 ml contains 70 gr of glucose.
   x          contains 20 gr of glucose.

x=1000*20/70
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10 days ago
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 [2617]
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
1 month ago
Calculate the molarity of 48.0 mL of 6.00 M H2SO4 diluted to 0.250 L
lorasvet [2549]

Answer:

The molality is 1.15 m.

Molality is calculated by dividing the number of moles of solute by the kilograms of solvent, which in this case is water.

Calculate moles of H₂SO₄ from molarity:

C = n/V → n = C × V = 6.00 mol/L × 0.048 L = 0.288 moles

Mass of solvent (water) based on density:

m = ρ × V = 1.00 kg/L × 0.250 L = 0.250 kg

Therefore, molality is:

m = moles/solvent mass = 0.288 moles / 0.250 kg = 1.15 m

4 0
1 month ago
Read 2 more answers
Describe the cause of attraction between molecules of water
lions [2669]
In a water molecule, the sharing of electrons occurs between the oxygen and hydrogen atoms within covalent bonds; however, this sharing is unequal. The oxygen atom holds a stronger pull on the electrons compared to the hydrogen atoms in the bond.
5 0
1 month ago
(a) The original value of the reaction quotient, Qc, for the reaction of H2(g) and I2(g) to form HI(g) (before any reactions tak
KiRa [2731]

Response:

Here's my calculation

Clarification:

Assume the starting concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.

We need to determine the initial concentration of HI.

1. We will need a chemical equation with concentrations, so let's compile all the information in one location.

H₂ + I₂ ⇌ 2HI

I/mol·L⁻¹: 0.30 0.15 x

2. Calculate the concentration of HI

Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} =  5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}

3. Plot the initial values

The graph below visualizes the initial concentrations as plotted on the vertical axis.

7 0
1 month ago
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