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How many grams of ammonia are necessary to form 9.09X^23 molecules of water?

User Morgler
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2 Answers

6 votes

Answer:

255 grams of ammonia

Step-by-step explanation:

To solve this problem, we need to use the balanced chemical equation for the reaction between ammonia and water:

NH3 + H2O → NH4+ + OH-

From this equation, we can see that one molecule of ammonia reacts with one molecule of water to produce one hydroxide ion (OH-) and one ammonium ion (NH4+). Therefore, we need the same number of molecules of ammonia as water to form the products.

So, if we have 9.09X^23 molecules of water, we need the same number of molecules of ammonia:

9.09X^23 molecules of NH3

To calculate the mass of ammonia required, we need to use the molar mass of ammonia, which is approximately 17 g/mol:

1 mol of NH3 = 17 g

To convert the number of molecules of NH3 to grams, we need to use Avogadro's number:

1 mol = 6.022 × 10^23 molecules

Therefore, the mass of ammonia required is:

9.09X^23 molecules of NH3 * (1 mol/6.022 × 10^23 molecules) * 17 g/mol

= 2.55 × 10^2 g or 255 grams (rounded to two significant figures)

So, we need 255 grams of ammonia to form 9.09X^23 molecules of water.

User Yancy
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7 votes

Answer:

I forgot dude sorry so sorry

User Jeevashankar
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