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nirvana33 [79]
3 years ago
12

Which substance is least soluble at 90°C?

Chemistry
1 answer:
jekas [21]3 years ago
7 0

Answer:

NH3

Explanation:

Which salt is least soluble at 90 degrees Celsius

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Will mark brainliest
Darina [25.2K]

Answer:

no because it can dissolve more solute at that temperature ...

6 0
4 years ago
the details on p. 99, in of wolves and men that describes how the wolf's fur carries seeds to depersed along his trail reveals..
Ymorist [56]

Answer is: the author’s perspective on events.

The book Of Wolfes and Men was written in 1978 by Barry Holstun Lopez.

Barry Lopez (1945 -) is an American author, essayist and fiction writer.

On this page, the wolf is not noticing the seeds on his fur. Seeds fall off as the wolf moves along the trail, so only author can make this observation.

6 0
3 years ago
When a hot piece of iron is placed in cool water,
Papessa [141]
A i believe is the answer
8 0
3 years ago
In chemistry lab student is determining the salinity of seawater using a conductivity meter. The student made standards of NaCl
dimaraw [331]

Answer:

a. is an excel document. i uploaded as an attachment

b. concentration = 0.327

Explanation:

in the graph i uploaded for answer a, the x variable is conductivity,  while the y variable is % salinity.

the calculated regression line was calculated using excel =

y = 0.057655X - 0.00891

for answer part b,

we take x = 5.82

when we put this into the y formular

y = 0.057655(5.82) - 0.00891

y = 0.3355521 - 0.00891

y = 0.3266421

this is ≈ 0.327

in conclusion, using the standard curve, the concentration of unknown salt is 0.327. the % salinity = 0.327

6 0
3 years ago
How many molecules of XeF6 are formed from 12.9 L of F2 (at 298 K and 2.6 atm) according to 11) the following reaction? Assume t
ddd [48]

Answer:

#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.

Explanation:

Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆

1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.

=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP

2. Calculate moles of F₂ used

=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used

3. Calculate moles of XeF₆ produced from reaction ratios …

Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆

4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number  (6.02 x 10²³ molecules/mole)

=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)

                                  = 2.75 x 10²³ molecules XeF₆.

8 0
3 years ago
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