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kondor19780726 [428]

3 years ago

11

1. If 3.2 L of nitrogen monoxide and 4.3 g of oxygen gas are combined, what volume of nitrogen dioxide will be formed? 2NO + 0,

-> 2NO,

1 answer:

Free_Kalibri [48]3 years ago

3 0

Answer:

$3.2 \div 22.4 = \frac{1}{7} mol \: no$

$4.3 \div 32 = 0.134mol \: o2$

Explanation:

NO is limit reactant so:

1÷7×46=6.57 grams of NO2 will form

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A certain alcohol contains only three elements, carbon, hydrogen, and oxygen. Combustion of a 60.00 gram sample of the alcohol p

marin [14]

Answer:

$C_2H_6O$

Explanation:

The first step is the <u>calculation of the moles</u> of $H_2O$ and $CO_2$ , so:

$114.6~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=2.6~mol~of~CO_2$

$70.44~g~H_2O\frac{1~mol~H_2O}{18~g~H_2O}=~3.9~mol~H_2O$

Now, in 1 mol of CO2 we have 1 mol of C and in 1 mol of $H_2O$ we have 1 mol of H. Additionally, if we want to calculate the moles of oxygen we need to <u>calculate the grams of C and O</u> and then do the <u>substraction</u> form the initial amount, so:

$2.6~mol~CO_2\frac{1~mol~C}{1~mol~CO_2}\frac{12~g~C}{1~mol~C}=31.25~g~of~C$

$3.9~mol~H_2O\frac{2~mol~H}{1~mol~H_2O}\frac{1~g~H}{1~mol~H}=7.82~g~of~H$

$Total~grams=~31.25~+~7.82=39.08~g$

$grams~of~O=60.00~g-~39.08~g=20.92~g~of~O$

Now we can <u>convert the grams</u> of O to moles, so:

$20.92~g~of~O\frac{1~mol~O}{16~g~O}=1.30~mol~O$

The next step is to divide all the mol values by the <u>smallest one</u>:

$O=\frac{1.30~mol~O}{1.30~mol~O}=~1$

$C=\frac{2.6~mol~C}{1.30~mol~O}=~2$

$H=\frac{7.82~mol~H}{1.30~mol~O}=6$

Therefore the formula is $C_2H_6O$

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If 100mL was diluted of a 2.0 M solution of sodium chloride by adding an additional 100 mL of water, what would the new molarity

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Tollens’ test, also known as silver-mirror test, is a qualitative laboratory test used to distinguish between an aldehyde and a

Makovka662 [10]

Explanation:

Tollens' reagent is prepared by using two-step process : -

Step 1:

Silver oxide is formed by mixing aqueous silver nitrate with base like sodium hydroxide. The reaction is shown below as:

$AgNO_3 + NaOH &\rightarrow AgOH + NHO_3 \nonumber \\ 2AgOH &\rightarrow Ag_2O + H_2O$

Step 2

Ammonia solution is drop-wise added until all the silver oxide dissolves to form the reagent. The reaction is shown below as:

$Ag_2O + 4NH_3 + H_2O \rightarrow 2Ag(NH_3)_2^+ + 2OH^-$

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Answer:

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