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lord [1]
3 years ago
13

1. El sorbitol, utilizado como edulcorante en algunos alimentos "sin azúcar" tiene un peso molecular de 182 g/mol y una composic

ión porcentual en masa de 39,56% de Carbono, 7,74% de Hidrógeno y 52,70% de Oxígeno. ¿Cuál es la fórmula molecular del Sorbitol? 2.- Se determina que una muestra de benzoato de metilo, un compuesto empleado en la elaboración de perfumes, contiene 70,57% de carbono, 5,93% de hidrógeno y 23,49% de oxígeno. Obtenga la fórmula molecular de esta sustancia si su peso molecular es de 136,1 g/mol. 3.- El etilenglicol, la sustancia empleada en los anticongelantes para automóvil, se compone de 38.7% en masa de Carbono, 9.7% en masa de Hidrógeno y 51.6% en masa de Oxigeno. Su masa molar es de 62.1 g/mol. Determine la fórmula molecular. 4.- El compuesto paradicloro se empleó a menudo como bola de naftalina. Si su análisis es 49.02% de C, 2.743% de H, y 48.24% de Cl, y su masa molecular, 147.0 g/mol. ¿Cuál es su fórmula molecular?.
Chemistry
1 answer:
Nostrana [21]3 years ago
5 0

Answer:

nnnnaaaoooo. seeeeiiii!!!!!!b k ibhhgj

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The liquid dispensed from a burette is called ___________.
11Alexandr11 [23.1K]

The liquid that is been dispensed during titration as regards this question is Titrant.

  • Titration can be regarded as  common laboratory method that is been carried out during quantitative chemical analysis.
  • This analysis helps to know the concentration of an identified analyte.
  • Burette can be regarded as laboratory apparatus.

It is used in the in measurements of variable amounts of liquid ,this apparatus helps in dispensation   of liquid, especially when performing titration.

  • The specifications is been done base on their volume, or resolution.
  • The liquid that comes out of this apparatus is regarded as Titrant, and this is gotten during titration process, which is usually carried out during volumetric analysis.

Therefore, burrete is used in volumetric analysis.

Learn more at:

brainly.com/question/2728613?referrer=searchResults

8 0
3 years ago
Read 2 more answers
What is the molar mass of a gas if 0.281 g of the gas occupies a volume of 125 ml at a temperature 126 °c and a pressure of 777
AleksAgata [21]
<em>Answer :</em> 72.05 g/mol
<span>
<em>Explanation : </em>

Let's </span>assume that the given gas is an ideal gas. Then we can use ideal gas equation,<span>
PV = nRT<span>
</span>
Where, 
P = Pressure of the gas (Pa)
V = volume of the gas (m³)
n = number of moles (mol)
R = Universal gas constant (8.314 J mol</span>⁻¹ K⁻¹)<span>
T = temperature in Kelvin (K)
<span>
The given data for the gas </span></span>is,<span>
P = 777 torr = 103591 Pa
V = </span>125 mL = 125 x 10⁻⁶ m³<span>
T = (</span>126 + 273<span>) = 399 K
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
n = ?

By applying the formula,
103591 Pa x  </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 399 K<span>
                                          n = 3.90 x 10</span>⁻³<span> mol

</span>Moles (mol) = mass (g) / molar mass (g/mol)<span>

Mass of the gas = </span><span>0.281 g
</span>Moles of the gas = 3.90 x 10⁻³ mol
<span>Hence,
   molar mass of the gas = mass / moles
                                          = 0.281 g / </span>3.90 x 10⁻³ mol
<span>                                          = 72.05 g/mol

</span>
7 0
3 years ago
Read 2 more answers
How many millimeters are in 251 cm?
antoniya [11.8K]

Answer:

2510

Explanation:

Since there are 10 milimetres in 1 centimetre, 1 cm = 10mm

There are 10 times as many milimetres as centimetres.

251cm X 10 = mm

251cm = 2510mm

3 0
3 years ago
Read 2 more answers
. In a separate experiment, the molar mass of nicotine is found to be somewhere between 150 and 180 g/mol. Calculate the molar m
stealth61 [152]

Answer:

<h2>         162g/mol</h2>

Explanation:

The question is incomplete. The complete question includes the information to find the empirical formula of nicotine:

<em>Nicotine has the formula   </em>C_xH_yN_z<em> . To determine its composition, a sample is burned in excess oxygen, producing the following results:</em>

  • <em>1.0 mol of CO₂</em>
  • <em>0.70 mol of H₂O</em>
  • <em>0.20 mol of NO₂</em>

<em>Assume that all the atoms in nicotine are present as products </em>

<h2>Solution</h2>

To find the empirical formula you need to find the moles of C, H, and N in each of the compound.

  • 1.0 mol of CO₂ has 1.0 mol of C
  • 0.70 mol of H₂O has 1.4 mol of H
  • 0.20 mol of NO₂ has 0.20 mol of N

Thus, the ratio of moles is:

  • C: 1.0
  • H: 1.4
  • N: 0.20

Divide all by the smallest number: 0.20

  • C: 1.0 / 0.20 = 5
  • H: 1.4 / 0.20 = 7
  • N: 0.20 / 0.20 = 1

Hence, the empirical formula is C₅H₇N

Find the mass of 1 mole of units of the empirical formula:

  • C:  5mol  × 12g/mol = 60g
  • H: 7mol × 1g/mol = 7 g
  • N: 1 mol × 14g/mol = 14g

Total mass = 60g + 7g + 14g = 81g

Two moles of units of the empirical formula weighs 2 × 81g = 162g and three units weighs 3 × 81g = 243 g.

Thus, since the molar mass is between 150 and 180 g/mol, the correct molar mass is 162g/mol and the molecular formula is twice the empirical formula: C₁₀H₁₄N₂.

5 0
3 years ago
Pls help me whit this pls
WITCHER [35]

Answer:

all cell arise from pre existing cell

Explanation:

6 0
3 years ago
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