Answer:
The solubility of X in water at 17°C is 0.110 g/mL.
Explanation:
The water of a rock pool lined with mineral crystals is a <em>saturated solution</em> of said mineral, this means the concentration of X in those 36 mL is the solubility of compound X in water at 17 °C.
- This means<u> it is possible to calculate said solubility</u>.
The dilution of the sample is not relevant, nor is that 500 mL volume. What's important is that 3.96 g of X form a saturated solution with 36.0 mL of water, so the solubility is:
- 3.96 g / 36.0 mL = 0.110 g/mL
<span>1. MgBr2
Soluble.
Rule: all the binary compounds of the group 17 (different to F) with metals are solubles, except those formed with Ag, Hg (I) and Pb.
2. PbI2
Insoluble.
Rule: it is one of the exceptions stated in the rule above.
3. (NH4)2CO3
Soluble.
Rule: salts containing NH4(+) are soluble.
4. ZnSO4
Soluble
Rule: </span><span>This salt is not an exception to the rule that most sulfate salts are soluble. Important exceptions to this rule include BaSO4,
PbSO4, Ag2SO4 and SrSO4
5. Sr(OH)2
Soluble (slightly soluble).
Rule: </span><span>Hydroxide salts of Group II elements (Ca,
Sr, and Ba) are slightly soluble</span>
Answer:
The first high part is Q4, then the low part is Q7, the following high part is Q6, and the energy moving from the next two high points is Q5.
Explanation:
The first high part is Q4, then the low part is Q7, the following high part is Q6, and the energy moving from the next two high points is Q5 because of the diagram.
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Answer:
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Explanation:
nice question...
ok solution goes right the way through...
in temp 193°C the volume of gas was 892ml
so for...144°C the volume of gas will be 892/193x144=665.533 or 665.534...(rounded off)...or 665°...(approx)....so got ur answer...
I hope it helps....
please comment if you have any further questions...i will be solving it for u...
please mark my answer brainliest