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Shtirlitz [24]
3 years ago
14

How many miles of NaBr will be produced when 3.33 mol of bromine reacts according to the following equation

Chemistry
1 answer:
serious [3.7K]3 years ago
7 0
Is there supposed to be more to this question? Maybe I can help.
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Wes melted some cheese over some tortilla chips to make nachos.
natima [27]

Answer:

a

Explanation:

Because the mass would stay the same and the flavor did not change and neither did the color so the shape is the only thing that changed

4 0
2 years ago
What is the identity of the element which had the following electron configuration? 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5 ​
kati45 [8]

Answer:

The element with electron configuration 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5 ​ is manganese (25Mn).

Explanation:

Step 1: Data given

The element with electron configuration 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5 ​

has 25 electrons.

This element has 2 electrons on the first shell, 8 electrons on the second shell, 13 electrons on the third shell and 2 electrons on the outer shell (valence electrons).

This means this element is part of group VII.

The element with 25 electrons, we can find on the periodic table, with atomic number 25.

The element with electron configuration 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5 ​ is manganese (25Mn).

5 0
3 years ago
How many atoms make up a diatomic molecule?
zalisa [80]
The prefix 'di' means two. Hence two atoms make up a diatomic molecule.
Hope this helps!
6 0
3 years ago
Encountering problems during separation/purification is a very common issue in organic synthesis, but these steps are crucial to
Rus_ich [418]
True is the correct answer! your welcome! :)
8 0
2 years ago
Given the following data:
bagirrra123 [75]

176.0 \; \text{kJ} \cdot \text{mol}^{-1}

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its \Delta H can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with \Delta H given be denoted as (1), (2), (3), and the last equation (4). Let a, b, and c be letters such that a \times (1) + b \times (2) + c \times (3) = (4). This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, 3 + (-1) = 2 shall resemble the number of \text{H}_2 left on the product side when the second equation is directly added to the third. Similarly

  • \text{NH}_4 \text{Cl} \; (s): -2 \; a = 1
  • \text{NH}_3\; (g): -2 \; b = -1
  • \text{HCl} \; (g): 2 \; c = -1

Thus

a = -1/2\\b = 1/2\\c = -1/2 and

-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)

Verify this conclusion against a fourth species involved- \text{N}_2 \; (g) for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

a + b = -1/2 + 1/2 = 0

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}

3 0
3 years ago
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