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prohojiy [21]
1 year ago
14

Identify the molecule with the highest boiling point. A(.C 6H 6 B(. NH 3 C(. OF 2 D(. CHCl 3

Chemistry
1 answer:
MaRussiya [10]1 year ago
7 0

Answer:

<em><u>C6H6</u></em>

Explanation:

  1. benzene (C6H6) is 80.1°C.
  2. ammonia (NH3) is -33.3°C
  3. oxygen difluoride (OF2) is -144.7°C.
  4. chloroform (CHCI 3) is 61.2°C.
You might be interested in
The density of silver is 10.5 g/cm3. What is the mass (in kilograms) of a cube of silver that measures 0.94 m on each side?
Rzqust [24]

Answer:

8721.132 Kg.

Explanation:

The following data were obtained from the question:

Density (D) of silver = 10.5 g/cm³

Length (L) of silver = 0.94 m

Mass (m) of silver =.?

Next, we shall determine the volume of silver. This can be obtained as follow:

Volume = Length × Length × Length

Volume = L × L × L

Volume = L³

Length (L) of silver = 0.94 m

Volume (V) of silver =?

Volume (V) of silver = 0.94³

Volume (V) of silver = 0.830584 m³

Next, we shall convert 0.830584 m³ to cm³. This can be obtained as follow:

1 m³ = 1×10⁶ cm³

Therefore,

0.830584 m³ = 0.830584 m³ / 1 m³ × 1×10⁶ cm³

0.830584 m³ = 830584 cm³

Therefore, 0.830584 m³ is equivalent 830584 cm³.

Thus, the volume of the silver is 830584 cm³.

Next, we shall determine the mass of the silver. This can be obtained as follow:

Density (D) of silver = 10.5 g/cm³

Volume of the silver = 830584 cm³.

Mass of silver =.?

Density = mass /

10.5 = mass of silver /830584

Cross multiply

Mass of silver = 10.5 × 830584

Mass of silver = 8721132 g

Finally, we shall convert the mass of silver, 8721132 g to kilogram (kg). This can be obtained as follow:

1000 g = 1 Kg

Therefore,

8721132 g = 8721132 g / 1000 g × 1 Kg

8721132 g = 8721.132 Kg

Therefore, the mass of each cube of silver is 8721.132 Kg

3 0
3 years ago
Fuel is injected at which state diesel cycle
frez [133]
The Diesel cycle<span> is a combustion process of a reciprocating </span>internal combustion in the cylinders of the engine<span>. In it, </span>fuel<span> is ignited by heat generated during the compression of air in the combustion chamber, into which fuel is then injected and the exhaust from the combustion is pushed out the exhaust pipe or brought out by a turbo </span>
6 0
3 years ago
PLEASE HELP- I WILL GIVE BRAINLIEST AND THANKS
julsineya [31]

Answer:

The mass of 10 cm³of a 0.4 g/dm³ solution of sodium carbonate is 0.004 grams

Explanation:

The question is with regards to density calculations

The density of the given sodium carbonate solution, ρ = 0.4 g/dm³

The volume of the given solution of sodium carbonate, V = 10 cm³ = 0.01 dm³

Density \ of \ an \ object, \rho  = \dfrac{The \ mass \ of \ the \ object, \ m }{\ The \ volume \ of \ the \ object, \ V }

\rho = \dfrac{m}{V}

Therefore, we have;

The \ Density \ of \ the \ sodium \ carbonate, \ \rho  = 0.4 \ g/dm^3 =  \dfrac{m }{ 0.01 \ dm^3 }

The mass, "m", of the sodium carbonate in  = ρ×V = 0.4 g/dm³ × 0.01 dm³ = 0.004 g

The mass of 10 cm³ (10 cm³ = 0.01 dm³) of a 0.4 g/dm³ solution of sodium carbonate, m = 0.004 g.

8 0
2 years ago
GIVING BRAINLY !!!!!! Calculate the molar mass of MgBr2. “SHOW” Your work!
Dovator [93]

Answer:

184.113 g/mol

Explanation:  The atomic mass of Mg is 24.3 amu. The atomic mass of bromine is 79.9. Therefore, the formula weight of MgBr2 equals 24.3 amu + (2 × 79.9 amu), or 184.1 amu. Because a substance's molar mass has the same numerical value as its formula weight, the molar mass of MgBr2 equals 184.1 g/mol.

8 0
2 years ago
PLEASE HELP ME!!!
kondor19780726 [428]
<h3>Answer:</h3>

0.424 J/g °C

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Chemistry</u>

<u>Thermochemistry</u>

Specific Heat Formula: q = mcΔT

  • q is heat (in Joules)
  • m is mass (in grams)
  • c is specific heat (in J/g °C)
  • ΔT is change in temperature
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] m = 38.8 g

[Given] q = 181 J

[Given] ΔT = 36.0 °C - 25.0 °C = 11.0 °C

[Solve] c

<u>Step 2: Solve for Specific Heat</u>

  1. Substitute in variables [Specific Heat Formula]:                                             181 J = (38.8 g)c(11.0 °C)
  2. Multiply:                                                                                                             181 J = (426.8 g °C)c
  3. [Division Property of Equality] Isolate <em>c</em>:                                                         0.424086 J/g °C = c
  4. Rewrite:                                                                                                             c = 0.424086 J/g °C

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.424086 J/g °C ≈ 0.424 J/g °C

6 0
2 years ago
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