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lozanna [386]
3 years ago
6

4. In a chromatography experiment, a student calculated that the solvent front was 6.8 cm above the starting line. Arginine trav

eled a distance of 4.9 cm and glycine traveled 3.4 cm. What Rf values would the student calculate for arginine and glycine
Chemistry
1 answer:
ziro4ka [17]3 years ago
3 0

Answer: 0.72 AND 0.5 RESPECTIVELY/.

Explanation:

The Rf value is defined as the ratio of the distance moved by the solute (i.e. the dye or pigment under observation) and the distance moved by the the solvent ie the migration of the solvent along the paper, where both distances are measured from point where the sample is first spotted on the paper.

Rf  value= (migration distance of substance)/(migration distance of solvent front)

or

Rf  value=Distance from Baseline travelled by Solute /Distance from Baseline travelled by Solvent (Solvent Front)

For Arginine Rf = (4.9 cm)/(6.8 cm) = 0.72

For glycine Rf = (3.4 cm)/(6.8 cm) = 0.5

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Bezzdna [24]

Answer:

\text{C. } _{36}^{85}\text{Kr}

Explanation:

Your nuclear equation is  

_{35}^{85}\text{Br} \longrightarrow \, _{-1}^{0}\text{e} +\, _{x}^{y}\text{X}

The main point to remember in balancing nuclear equations is that

  • the sum of the superscripts and must be the same on each side of the equation.
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Then  

85 = 0 + y, so y = 85 - 0 = 0  

35 = -1 + x, so x = 35 + 1 = 36

The nucleus with atomic number 36 and atomic mass 85 is krypton-85.  

The nuclear equation becomes  

_{35}^{85}\text{Br} \longrightarrow \, _{-1}^{0}\text{e} + \, _{36}^{85}\text{Kr}

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3 years ago
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If 3.75 moles of aluminum react, how many moles of sulfur are needed?
aliya0001 [1]

Answer:(A) 0.70 mol

Explanation:

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otez555 [7]

Answer:

1. 2Ca + N_{2} → 2CaN

2. 4Li + O_{2} → 2Li_{2}O

3. 2KCl + BaF_2 → 2KF + BaCl_2

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What is the freezing point of a solution made with 1.31 mol of CHCl3 in 530.0 g of CCl4 (Kf =29.8 degrees C/m, Freezing point of
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73.606 °C is the freezing point of the solution made with with 1.31 mol of CHCl3 in 530.0 g of CCl4.

Explanation:

Data given:

number of moles of CHCl3 = 1.31 moles

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Kf = 29.8 degrees C/m

freezing point of pure solvent or CCl4 =  -22.9 degrees

freezing point = ?

The formula used to calculate the freezing point of the mixture is

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molality = \frac{moles of solute}{mass of solvent in kilograms}

putting the value in the equation:

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ΔT = 1.31 x 29.8 x 2.47

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