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1 year ago

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An unsaturated aqueous solution of NH3 is at 90.°C in 100. grams of water. According to Reference Table G, how many grams of NH3

could this unsaturated solution contain?
(1) 5 g (3) 15 g
(2) 10. g (4) 20. g

1 answer:

Illusion [34]1 year ago

3 0

The answer will be 2. 10.g
Hope it's help you

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What is the index of hydrogen deficiency of a compound with a molecular formula of c6h15n?

7nadin3 [17]

Answer:

IHD = 0

Explanation:

Given that

C₆H₁₅N

Number of carbon atoms(n) = 6

Number of hydrogen atoms(x') = 15

Number of nitrogen atoms = 1

There is nitrogen atoms then x = x' -1

The index of hydrogen deficiency given as

$IHD=\dfrac{2n+2-x}{2}$

So

$IHD=\dfrac{2n+2-x}{2}$

$IHD=\dfrac{2\time 6+2-(15-1)}{2}$

IHD = 0

The index of hydrogen deficiency is zero.

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2 years ago

That's just the tip of the iceberg" is a popular expression you may have heard. It means that what you can see is only a small p

vitfil [10]

<span>The density of an iceberg is less than that of water and that is why it floats. It is the same as ice cubes floating on water too. Water is a very unique substance in that it is one of few compounds where cooling it past freezing point decreases its density (study hydrogen bonds). The possible answers are therefore A or C. If the majority of the ice is below the water then it should be clear from common sense that A is the correct answer as it would mostly float on the top with just a little under the surface if the answer were as low as C. See Archimedes Principle for an explanation of how much of the ice floats and how much is underwater.</span>

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1 year ago

Read 2 more answers

At the beginning of the school year, a chalk company receives an order for 2000 boxes which is the largest order ever placed. Ea

lys-0071 [83]

Answer:

<u>259,000 g of chalk.</u>

Explanation:

<u>1) Data:</u>

a) 2000 boxes

b) 175 g / box

c) % yield = 74%

<u>2) Formula: </u>

% yield = (theoretical yield / actual yield) × 100

<u>3) Solution:</u>

a) Calcualte the actual yield:

mass of product = 2000 box × 175 g/ box = 350,000 g

b) Solve for the theoretical yield from the % yield formula:

% yield = (theoretical yield / actual yield) × 100

⇒ theoretical yield = % yield × actual yield / 100

theoretical yield = 74% × 350,000g / 100 = 259,000 g

6 0

1 year ago

Octane (C8H18) undergoes combustion according to the following thermochemical equation. 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l

Zepler [3.9K]

Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol

Explanation:

The given balanced chemical reaction is,

$2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(l)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

We are given:

$\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_8H_{18}(l))}=?kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol$

Putting values in above equation, we get:

$-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]$

$\Delta H^o_f_{(C_8H_{18}(l))}=-250.2kJ/mol$

Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol

4 0

2 years ago

Calculate the radius ratio for NaBr if the ionic radii of Na + and Br − are 102 pm and 196 pm , respectively. radius ratio: Base

Fudgin [204]

Answer : The expected coordination number of NaBr is, 6.

Explanation :

Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.

This is represented by,

$\frac{r_{cation}}{r_{anion}}$

When the radius ratio is greater than 0.155, then the compound will be stable.

Now we have to determine the radius ration for NaBr.

Given:

Radius of cation, $Na^+$ = 102 pm

Radius of cation, $Br^-$ = 196 pm

$\frac{r_{cation}}{r_{anion}}=\frac{102}{196}=0.520$

As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.

The relation between radius ratio and coordination number are shown below.

Therefore, the expected coordination number of NaBr is, 6.

8 0

1 year ago