MgBr2(aq) is an ionic compound which will have the releasing of 2 Br⁻ ions ions in water for every molecule of MgBr2 that dissolves.
MgBr2(s) --> Mg+(aq) + 2 Br⁻(aq)
[Br⁻] = 0.51 mol MgBr2/1L × 2 mol Br⁻ / 1 mol MgBr2 = 1.0 M
The answer to this question is [Br⁻] = 1.0 M
Answer:
Compared to other stars we observe, the Sun appears bigger and brighter because it is much closer to earth.
Objects closer to earth appear to be much bigger and brighter.
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Answer:
7.5 g
Explanation:
There is some info missing. I think this is the original question.
<em>Ammonium phosphate ((NH₄)₃PO₄) is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid (H₃PO₄) with ammonia (NH₃). What mass of ammonium phosphate is produced by the reaction of 4.9 g of phosphoric acid? Be sure your answer has the correct number of significant digits.</em>
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Step 1: Write the balanced equation
H₃PO₄ + 3 NH₃ ⇒ (NH₄)₃PO₄
Step 2: Calculate the moles corresponding to 4.9 g of phosphoric acid
The molar mass of phosphoric acid is 98.00 g/mol.
Step 3: Calculate the moles of ammonium phosphate produced from 0.050 moles of phosphoric acid
The molar ratio of H₃PO₄ to (NH₄)₃PO₄ is 1:1. The moles of (NH₄)₃PO₄ produced are 1/1 × 0.050 mol = 0.050 mol.
Step 4: Calculate the mass corresponding to 0.050 moles of ammonium phosphate
The molar mass of ammonium phosphate is 149.09 g/mol.
The s orbitals are not symmetrical in shape is a FALSE statement.
An s orbital is so symmetric, more specifically spherically symmetric that it looks the same from all directions.
- The atomic orbitals in the atoms of elements differ in shape.
In essence, the electrons they describe have varying probability distributions around the nucleus. The spherical symmetry of s orbitals is evident in the fact that all orbitals of a given shell in the hydrogen atom have the same energy.
- All s orbitals are spherically symmetrical. Put simply, an electron that occupies an s orbital can be found with the same probability at any orientation (at a distance) from the nucleus.
The s orbitals are therefore represented by a spherical boundary surface which is a surface which captures a high proportion of the electron density.
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