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Dimas [21]
4 years ago
11

What medal is liquid at room temperature?

Chemistry
2 answers:
jok3333 [9.3K]4 years ago
5 0
Gallium is a metal that is a liquid at room temperature.
DIA [1.3K]4 years ago
5 0
I do believe it is Gallium. I heard that it melts quickly to liquid by just sitting on your hand! (cool)
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Ferric chloride is being applied for coagulation at a 15.0 MGD surface water treatment plant, at a rate of 195 pounds per day. A
s344n2d4d5 [400]

5850 pounds of this chemical will be used in a 30-day month.

195 pounds per day.

So, 30 days = 195 * 30

= 5850 pounds

Iron(III) chloride is the inorganic compound with the components FeCl 3. additionally known as ferric chloride, it is a commonplace compound of iron inside the +3 oxidation country. The anhydrous compound is crystalline and stable with a melting point of 307.6 °C.

The primary use of ferric chloride is to dispose of impurities in water and for wastewater treatment. Ferric chloride is likewise one of the few water remedy chemical compounds that can sequester odors.

Reasonably poisonous with the aid of ingestion but a strong irritant to the skin and eyes. it's far moderately corrosive and skin touch should be avoided and avoid skin and eye contact.

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6 0
2 years ago
A student performs an experiment to determine the density of a sugar solution. She obtains the following results: 1.11 g/mL, 1.8
ehidna [41]

Answer:

D. her results are neither precise, nor accurate

Explanation:

Reason being that: First, she obtained four different result for an experiment ( obviously it's supposed to be one value for the different procedure or atleast very close range) sugar solution and all four values were far apart.

6 0
3 years ago
Which of the following materials would be most easily scratched? 1.Gypsum
Effectus [21]
The answer is quartz where the hardness is 7.

Can scratch glass easily. This is where most granites are rated in terms of hardness
5 0
3 years ago
Read 2 more answers
URGENT!...
exis [7]

Answer: C. There is more plate movement in California.

Explanation:

4 0
4 years ago
Suppose 10.0 mL of 2.00 MNaOH is added to (a) 0.780 L of pure water and (b) 0.780 L of a buffer solution that is 0.682 Min butan
katrin2010 [14]

Answer:

a) pH will be 12.398

b) pH will be 4.82.

Explanation:

a) The moles of NaOH added = molarity X volume (L) = 2 X 0.01 = 0.02 moles

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The new concentration of /NaOH will be:

molarity=\frac{molesofsolute}{volumeofsolution}=\frac{0.02}{0.79}=0.025M

the [OH⁻] = 0.025

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pH = 14 -pOH = 12.398

b) The buffer has butanoic acid and butanoate ion.

i) Before addition of NaOH the pH will be calculated using Henderson Hassalbalch's equation:

pH=pKa+log\frac{[salt]}{[acid]}

pKa=-logKa=-log(1.5X10^{-5})=4.82

ii) on addition of base the pH will increase.

8 0
4 years ago
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