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UkoKoshka [18]
3 years ago
9

Given a stock standard solution of concentration 0.200 M, calculate the volume in mL required to make up calibration standards o

f the following concentrations (in 25.0 mL volumetric flasks)
1.60 x 10-2 M .....mL
Chemistry
1 answer:
guapka [62]3 years ago
6 0

M1V1 = M2V2

.200 (.025) = 1.60 X 10 -2 (V2)

V2 = .315 L

1.60 x 10-2 M  in 315 mL

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What are the 7 methods of separating mixtures?​
Gennadij [26K]

Methods Of Separating Mixtures

Handpicking.

Threshing.

Winnowing.

Sieving.

Evaporation.

Distillation.

Filtration or Sedimentation.

Separating Funnel.

3 0
3 years ago
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A magnesium ion, Mg2+, is formed from a magnesium atom, Mg.
konstantin123 [22]

Answer:

The correct row is B.

Explanation:

The formation of the magnesium ion, Mg²⁺, comes from the removal of 2 electrons from the valence shell of the magnesium atom.<em> </em>Since the remotion is of electrons, the nucleus of the atom remains the same, so the number of protons and the number of neutrons does not change.<em> </em>

If the number of proton change, then the atom also change, since the identity of an atom is related to the atomic number which is the same to the proton number.          

Now, if the number of neutrons changes, then we would be in the presence of an isotope of the magnesium atom.        

Therefore, the correct row is B.  

I hope it helps you!                            

5 0
3 years ago
Compare wave A with wave B correctly in relation to amplitude.
Anni [7]
<span>Wave A will have a higher pitch than wave B.. This is not true becasue b has the higher PITCH becasue it's closer together.

Wave B will have a lower pitch than wave A... This basically A but worded differently. 

Wave A will have a louder sound than wave B... This is CORRECT becasue it's louder the waver are bigger, so it's louder. But it has a lower pitch, 2 different things. 

Wave B will have a louder sound than wave A. False..</span><span />
7 0
3 years ago
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A sample of gas (1.9 mol) is in a flask at 21 °c and 697 mm hg. the flask is opened and more gas is added to the flask. the new
garik1379 [7]

To solve this problem, we assume ideal gas so that we can use the formula:

PV = nRT

since the volume of the flask is constant and R is universal gas constant, so we can say:

n1 T1 / P1 = n2 T2 / P2

 

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<span>n2 = 2.25 moles</span>

8 0
3 years ago
A 2.0 molal sugar solution has approximately the same freezing point as 1.0 molal solution of 1) CaCl2 2) CH3COOH 3) NaCl 4) C2H
sertanlavr [38]

Answer:

3) NaCl.

Explanation:

<em>∵ ΔTf = iKf.m</em>

where, <em>i</em> is the van 't Hoff factor.

<em>Kf </em>is the molal depression freezing constant.

<em>m</em> is the molality of the solute.

<em>The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. </em>

<em></em>

  • For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

<em>So, for sugar: i = 1.</em>

<em>∴ ΔTf for sugar = iKf.m = (1)(Kf)(2.0 m) = 2 Kf.</em>

<em></em>

  • For most ionic compounds dissolved in water, the van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance.

For NaCl, it is electrolyte compound which dissociates to Na⁺ and Cl⁻.

<em>So, i for NaCl = 2.</em>

<em>∴ ΔTf for NaCl = iKf.m = (2)(Kf)(1.0 m) = 2 Kf.</em>

<em></em>

<em>So, the right choice is: 3) NaCl.</em>

<em></em>

8 0
3 years ago
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