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2 years ago

NEED IT ASAP!PLEASE HURRY

Chemistry

1 answer:

marin [14]2 years ago

7 0

Answer:

snow, or freezing rain?

Explanation:

because freezing rain can occur then, but snow is more common.

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Which of the following refers to all of the organisms in a particular area?

Ivan

biological community is an interacting group of various species in a common location.

4 0

3 years ago

When water freezes,it contracts true or false

Zigmanuir [339]

The correct answer is: False.
_______________________________________________________
Water expands (gets larger/ is less dense in solid form than in liquid form) when it freezes).
_______________________________________________________

6 0

4 years ago

Read 2 more answers

Naturally occurring Indium has two isotopes.

trasher [3.6K]

Answer:

Average Molar mass is 114.8u

Explanation:

» From the formula of average Relative Atomic Mass [ RAM ]

${ \rm{RAM = \sum \frac{(isotopic \: mass \times \%abundance)}{100} }} \\$

» We have two isotopes of Indium,

Indium-113, mass » 112.9u, % » 4.28%
Indium-115, mass » 114.9u, % » 95.72%

${ \tt{RAM = \frac{ \blue{(112.9 \times 4.28)} + { \green{(114.9 \times 95.72)}}}{100} }} \\ \\ { \tt{RAM = \frac{11481.44}{100} }} \\ \\ { \underline{ \underline{ \tt{ \: RAM = 114.8\: }}}}$

3 0

2 years ago

If 13.2 kg of Al 2 O 3 ( s ) , 50.4 kg of NaOH ( l ) , and 50.4 kg of HF ( g ) react completely, how many kilograms of cryolite

Elodia [21]

Answer : The mass of cryolite produced will be, 54.38 kg

Solution : Given,

Mass of $Al_2O_3$ = 13.2 kg = 13200 g

Mass of $NaOH$ = 50.4 kg = 50400 g

Mass of $HF$ = 50.4 kg = 50400 g

Molar mass of $Al_2O_3$ = 101.9 g/mole

Molar mass of $NaOH$ = 40 g/mole

Molar mass of $HF$ = 20 g/mole

Molar mass of $Na_3AlF_6$ = 209.9 g/mole

First we have to calculate the moles of $Al_2O_3,NaOH$ and $HF$ .

$\text{ Moles of }Al_2O_3=\frac{\text{ Mass of }Al_2O_3}{\text{ Molar mass of }Al_2O_3}=\frac{13200g}{101.9g/mole}=129.54moles$

$\text{ Moles of }NaOH=\frac{\text{ Mass of }NaOH}{\text{ Molar mass of }NaOH}=\frac{50400g}{40g/mole}=1260moles$

$\text{ Moles of }HF=\frac{\text{ Mass of }HF}{\text{ Molar mass of }HF}=\frac{50400g}{20g/mole}=2520moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Al_2O_3+6NaOH+12HF\rightarrow 2Na_3AlF_6+9H_2O$

From the balanced reaction we conclude that

The mole ratio of $Al_2O_3,NaOH$ and $HF$ is, 1 : 6 : 12

And, the ratio of given moles of $Al_2O_3,NaOH$ and $HF$ is, 129.54 : 1260 : 2520

From this we conclude that, $NaOH,HF$ is an excess reagent because the given moles are greater than the required moles and $Al_2O_3$ is a limiting reagent and it limits the formation of product.