All categories

3 years ago

Which three elements are liquid at 25 degrees Celsius

Chemistry

2 answers:

KengaRu [80]3 years ago

5 0

Answer:

Bromine and mercury are liquid at room temperature and gallium is liquid when just a little above room temperature. Originally Answered: What are the only two elements that are liquid at 25° C (room temperature)? Mercury and Bromine. Mercury is solid below -38 degrees F (-39C) and Bromine below 19 degrees F (-7.2C).

Explanation:

Blababa [14]3 years ago

3 0

It’s bromine and mercury are liquid at room temperature I don’t really know about any other but another element like gallium starts melting at the same temperature or slightly higher

You might be interested in

How many grams of water at 50◦c must be added to 16 grams of ice at −12◦c to result in only liquid water at 0◦c?

melisa1 [442]

When water at 50 C is added to ice at -12 C, heat is transferred from hot water to ice.

- Heat given out by water = Heat absorbed by ice

Calculating the heat released by hot water:

$q = m Cwater$ ΔT $q = m (4.184\frac{J}{(g.^{0}C)} )$ $(0^{0}C-50^{0}C)$

Calculating heat absorbed by 16 g of ice: Ice at $-12^{0}C$ is converted to ice at $O^{0}C$ and then ice at $O^{0} C$ to water at $0^{0}C$

$q = m Cice$ ΔT + $m (Heat of fusion)$

$q = 16 g(2.11\frac{J}{g.^{0}C})(0^{0}C - (-12^{0}C))$ + $16 g (333.55\frac{J}{g})$

q = 405.12 J +5336.8 J =5741.92 J

- Heat given out by water = Heat absorbed by ice

-( $m(4.184\frac{J}{g.^{0}C})(0^{0}C-50^{0}C) = 5741.92 J$

m = 27.4 g

Therefore, 27.4 g water at $50^{0}C$ must be added to 16 g of ice at $-12^{0}C$ to convert to liquid water at $0^{0}C$

4 0

3 years ago

A mixture of two compounds, A and B, was separated by extraction. After the compounds were dried, their masses were found to be:

Arlecino [84]

Answer:

The percent recovery from re crystallization for both compounds A and B is 69.745 and 81.44 % respectively.

Explanation:

Mass of compound A in a mixture = 119 mg

Mass of compound A after re-crystallization = 83 mg

Percent recovery from re-crystallization :

$\frac{\text{Mass after re-crystallization}}{\text{Mass before re-crystallization}}\times 100$

Percent recovery of compound A:

$\frac{83 mg}{119 mg}\times 100=69.74\%$

Mass of compound B in a mixture = 97 mg

Mass of compound B after re-crystallization = 79 mg

Percent recovery of compound B:

$\frac{79 mg}{97 mg}\times 100=81.44\%$

3 0

3 years ago

3. How many molecules are in 0.500 moles of sulfur?

satela [25.4K]

Explanation:

No of molecules=0.500×6.023×10²³=3.011×10²³ molecules

7 0

3 years ago

You have 0.14 moles of sodium chloride (NaCl). How many grams do you have? ( 1 mole of NaCl = 58 grams NaCl) *

Step2247 [10]

Answer:

Option D is correct = 8.12 grams of NaCl

Explanation:

Given data:

Moles of sodium chloride = 0.14 mol

Mass of sodium chloride = ?

Solution:

Formula:

Number of moles = mass of NaCl / Molar mass of NaCl

Molar mass of NaCl = 58 g/mol

Now we will put the values in formula.

0.14 mol = Mass of NaCl / 58 g/mol

Mass of NaCl = 0.14 mol × 58 g/mol

Mass of NaCl = 8.12 g of NaCl

Thus, 0.14 moles of NaCl contain 8.12 g of NaCl.

4 0

3 years ago

What is the molar mass of BaBr2?<br><br>multiple choice on the picture

laiz [17]

B. 297.1 g/mol

Is your answer! ;)

3 0

3 years ago

Read 2 more answers