Answer:
substance A is a liquid, while substance B is a liquid or gas
Answer:

Explanation:
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In this case, since the percent water is computed by dividing the amount of water by the total mass of the hydrate; we infer we first need the molar mass of water and that of the hydrate as shown below:

Thus, the percent water is:

So we plug in to obtain:

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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>
<em />
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.

Answer : The concentration of NOBr after 95 s is, 0.013 M
Explanation :
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant =
t = time taken = 95 s
[A] = concentration of substance after time 't' = ?
= Initial concentration = 0.86 M
Now put all the given values in above equation, we get:
![0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)](https://tex.z-dn.net/?f=0.80%3D%5Cfrac%7B1%7D%7B95%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%280.86%29%7D%5Cright%29)
[A] = 0.013 M
Hence, the concentration of NOBr after 95 s is, 0.013 M