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OlgaM077
1 month ago
9

Determine the relative formula mass of hexasodium difluoride using the periodic table below. A. 138 g/mol B. 176 g/mol C. 20 g/m

ol D. 42 g/mol
Chemistry
2 answers:
VMariaS [2.9K]1 month ago
4 0

Answer:

B. 176 g/mol

Explanation:

chemistry ed tell

lions [2.9K]1 month ago
3 0

Answer:

Option B. 176 g/mol

Explanation:

Let's begin by establishing the chemical formula for hexasodium difluoride:

Hexasodium indicates there are 6 sodium atoms.

Difluoride signifies there are 2 fluorine atoms.

Consequently, the formula for hexasodium difluoride is Na6F2.

The relative formula mass of a compound is determined by adding the atomic weights of its elements together.

Thus, the relative formula mass for hexasodium difluoride, Na6F2, is calculated as follows:

Molar mass of Na = 23 g/mol

Molar mass of F = 19 g/mol

Relative formula mass Na6F2 = (23x6) + (19x2)

= 138 + 38

= 176 g/mol

Thus, the relative formula mass of hexasodium difluoride, Na6F2, equals 176 g/mol

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(CH3)2N2H2 + N2O4 → N2 + H2O + CO2 + heat [balance?]​
Anarel [2989]

Answer:

2(CH3)2N2H2 + 3N2O4 → 4N2 + 4H2O + 4CO2 + heat

Explanation:

  • To balance chemical equations, coefficients are assigned to both reactants and products.
  • This yields an equal count of atoms of each element on both sides of the equation.
  • Balancing chemical equations ensures compliance with the law of conservation of mass.
  • According to this law, the mass of reactants must equal the mass of products, achievable through balancing the equation.
  • The application of coefficients 2, 3, 4, 4, 4 allows for an equal balance in the equation.
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       2(CH3)2N2H2 + 3N2O4 → 4N2 + 4H2O + 4CO2 + heat

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18 days ago
Given the connection between Aw and K (Aw=2k) could you use the ideal gas law and derive the Boltzmann constant. Water freezes a
lions [2927]

Answer:

Explanation:

The relationship between the new temperature scale and the absolute temperature scale is defined as follows

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Aw = 2 x 273.15 = 546.3 K

Each division of the new scale is equivalent to half that of each division on the absolute scale

each division of the new scale is minimal.

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Here, per K corresponds to 2Aw

Hence, the value of R in the new scale = 8.314/2 J per mole per Aw

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= .69 x 10⁻²³

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7 0
2 months ago
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A student reported to her instructor that her unknown contained salt, salicylic acid, and sand. In reality the unknown contained
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Control: freezer and ice tray

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