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2 years ago

What is the weight of oxygen in 50gwater

Chemistry

1 answer:

8_murik_8 [283]2 years ago

5 0

To compute the required mass of the element to form the given molecules, we'll use the general formula shown below:

{eq}\displaystyle m_1=\frac{M_1}{M_2}\times m_2 {/eq}, where,

{eq}M_1 {/eq} is the molecular weight of the element.

{eq}M_2 {/eq} is the molecular weight of the given molecule.

{eq}m_2 {/eq} is the mass of the given molecule.

Answer and Explanation:

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The given value is:

The value of the mass of the water molecule is {eq}m_2=\rm 50\ grams {/eq}.

The chemical reaction for the formation of water...

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Formula for Mass: Definition & Examples

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Chapter 28 / Lesson 42

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What is mass? Learn the mass definition and how to find the mass of an object using the mass formula. See common mass characteristics and properties.

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A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percentage, the student measures out 5.84 grams o

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Answer:

The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml

Explanation:

Here we have the reaction of AgNO₃ and NaCl as follows;

AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)

Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,

Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains

$Number \, of \, moles, n = \frac{Mass}{Molar \ mass} = \frac{5.84}{58.44} = 0.09993 \ moles \ of \ NaCl$

0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃

Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;

0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL

Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.

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3 years ago

What is the weight of oxygen in 50gwater​

What is the weight of oxygen in 50gwater