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Diano4ka-milaya [45]
3 years ago
9

Which metal would most likely have the highest melting point?

Chemistry
2 answers:
AURORKA [14]3 years ago
5 0

Answer:

Tungsten

Explanation:

Anettt [7]3 years ago
4 0

Answer:

tungsten

Explanation:

I hope it's helpful for you

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Please answer this as soon as possible Thank You.
sashaice [31]

Answer:

jchb3prhoucheoucheoucheouchoudehocuehcdouhd

Explanation:

5 0
3 years ago
50 POINTS FOR THIS!
Leona [35]

first there is only 25 points on here because if you add 50 points to the question brainly divides it by two so that two people can answer so if you want each person to get 50 points or to advertise how much point sis on this quetstion you would instead need to add 100 points to the question- anyways let's answer this question now

Do you know what the Law of Conservation of mass says? Matter is neither created nor destroyed. Therefore, you have started with 12.7 g Cu and 3.2 g S on the left for a total of 15.9 g; therefore, you must obtain 15.9 g of product(s). If Cu2S is the only product there must be 15.9 g Cu2S.

5 0
3 years ago
Significant figures<br> 5.3316 + 6.87 + 37.48
Aleksandr-060686 [28]

Answer:

Explanation:

01: 5.3316+6.87+37.48

02: 12.2016+37.48

03: 49.6816

04: 49.68

Answer: 49.68 (Decimals: 2; Significant Figures: 4)

HOPE THIS HELPS..

5 0
4 years ago
Read 2 more answers
Which of the following correctly lists the name of the element, the symbol for its ion, and the name of that ion?
amm1812
B. Aluminum, Al3+, aluminum ion is correct

Chloride would be negative, Iron has Fe for the symbol, potassium is K....
3 0
4 years ago
Read 2 more answers
9
yaroslaw [1]

Answer:

T2 = 36.38°C

Explanation:

Given data:

Mass of water = 75 g

Initial temperature = 30 °C

Final temperature = ?

Heat absorbed = 2000 J

Solution:

Specific heat capacity of water is 4.18 J/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

2000 J = 75 g×4.18 J/g.°C × T2- T1

2000 J = 313.5 J/°C × T2- T1

2000 J = 313.5 J/°C × T2 - 30 °C

2000 J / 313.5 J/°C  = T2 - 30 °C

6.38 °C = T2 - 30 °C

T2 = 6.38 °C +  30°C

T2 = 36.38°C

7 0
3 years ago
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