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taurus [48]
3 years ago
10

What is the name of the compound with the formula B2C14?​

Chemistry
1 answer:
agasfer [191]3 years ago
6 0

Answer:

Diboron tetrachloride --->  B2Cl4

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You wish to make a 0.285 M hydroiodic acid solution from a stock solution of 12.0 M hydroiodic acid. How much concentrated acid
Inessa05 [86]

The amount, in mL, of the concentrated acid required, would be 1.1875 mL

<h3>Dilution</h3>

From the dilution equation:

m1v1=m2v2 where m1 and m2 = molarity before and after dilution, and v1 and v2 = volume before and after dilution.

m2 = 0.285M, m1 = 12.0M v2 = 50.0 mL

v1 = m2v2/m1 = 0.285x50/12 = 1.1875 mL

Thus, 1.1875 mL of the acid would be taken and diluted with water up to the 50 mL mark.

More on dilution can be found here: brainly.com/question/13949222

#SPJ1

4 0
2 years ago
How many meters does daniel have to walker
Tcecarenko [31]

14,200 because all you have to do to solve this is multiply 14.2 kilometers by 1,000 meters to find the distance that he walks.

8 0
3 years ago
Read 2 more answers
You mix a blue and a clear liquid together and they turn bright green. Has a chemical reaction occurred?
Margarita [4]
Yes because color change is a sign of a chemical reaction.
7 0
3 years ago
Read 2 more answers
A 0.5376 g sample of an unknown compound is found to contain 0.3044 g of carbonate. Could this compound be calcium carbonate?
shepuryov [24]
Calcium carbonate has the formula: CaCO3
From the periodic table:
mass of calcium = 40 grams
mass of carbon = 12 grams
mass of oxygen = 16 grams
Therefore,
molar mass of CaCO3 = 40 + 12 + 3(16) = 100 grams
molar mass of carbonate = 12 + 3(16) = 60 grams

One mole of calcium carbonate contains one mole of carbonate. Therefore, 100 grams of CaCO3 contains 60 grams of CO3.
If the 0.5376 grams of the unknown substance is CaCO3, then the amount of carbonate will be:
amount of carbonate = (0.5376*60) / 100 = 0.32256 grams

Based on the above calculations, the sample is not CaCO3
6 0
3 years ago
When 551. mg of a certain molecular compound X are dissolved in 100 g of benzonitrile (CH,CN), the freezing point of the solutio
arsen [322]

Answer:

1.12g/mol

Explanation:

The freezing point depression of a solvent for the addition of a solute follows the equation:

ΔT = Kf*m*i

<em>Where ΔT is change in temperature (Benzonitrile freezing point: -12.82°C; Freezing point solution: 13.4°C)</em>

<em>ΔT = 13.4°C - (-12.82) = 26.22°C</em>

<em>m is molality of the solution</em>

<em>Kf is freezing point depression constant of benzonitrile (5.35°Ckgmol⁻¹)</em>

<em>And i is Van't Hoff factor (1 for all solutes in benzonitrile)</em>

Replacing:

26.22°C = 5.35°Ckgmol⁻¹*m*1

4.90mol/kg = molality of the compound X

As the mass of the solvent is 100g = 0.100kg:

4.9mol/kg * 0.100kg = 0.490moles

There are 0.490 moles of X in 551mg = 0.551g, the molar mass (Ratio of grams and moles) is:

0.551g / 0.490mol

= 1.12g/mol

<em>This result has no sense but is the result by using the freezing point of the solution = 13.4°C. Has more sense a value of -13.4°C.</em>

5 0
2 years ago
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