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Firdavs [7]
3 years ago
11

Mercury is a liquid meteal but it is not hard? how can we say it is a metal

Chemistry
1 answer:
NNADVOKAT [17]3 years ago
7 0
What's the problem ? Hardness is not the definition of a metal. You need to expand your thinking. EVERY element is solid, liquid, and gas, over different ranges of temperature ... including all of the metals. There are only TWO elements that are liquid AT ROOM TEMPERATURE, and mercury is one of them. But on a mild day at the south pole, mercury is solid too.
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Difference between giant ionic structure,giant covalent structure and simple molecular structure and also covalent bonds and ion
Alex_Xolod [135]
Simple molecular biology
8 0
3 years ago
3.
belka [17]

Answer:

3. Inverse 1. Direct

Explanation:

P- pressure

V - volume

T - temperature

P1*V1 / T1 = P2*V2 / T2 ...... (1)

That's the general gas law with the combined ideas of charles, boyle & lussac.

Whenever you are restricted as "constant" temperature, volume, or pressure...cancel them off of your equation.

in this case 3. is indirectly telling us to cancel the temperature (T).

so we'll be left w P1*V1 = P2*V2

now notice that any relation ship that is multiplied like the one above consists of inversely related quantities. & so we conclude that-

P & V are inversely proportional or have an inverse relationship.

similarly in 1. we'll cancel p off of the general formula (1)

to be left with V1/T1 = V2/T2

also note that quantities involved in division are directly related to each other & hence the answer.

5 0
2 years ago
How much potassium nitrate could be dissolved into 2L of water
miv72 [106K]

Answer:

About 110 g.

Your tool of choice here will be the solubility graph for potassium nitrate, KNO3, in water.

8 0
3 years ago
Read 2 more answers
Current Question:
IRISSAK [1]

Answer:

0.2g

Explanation:

All radiodecay follows the 1st order decay equation

A = A₀e^-kt

A => Activity at time (t)

A₀ => Initial Activity at time = 0

k => decay constant for isotope

T => time in units that match the decay constant

Half-Life Equation => kt(½) = 0.693 => k = 0.693/34 min = 0.0204min¹

A = A₀e^-kt  = (26g)e^-(0.0204/min)(238min) = (26g)(0.0078) = 0.203g ~ 0.2g (1 sig fig).

7 0
3 years ago
The specific heat of copper metal is 0.385 J/(g × °C). How much energy must be added to a 35.0 gram sample of copper to change t
Montano1993 [528]
Q = mC∆T 

<span>where: </span>
<span>q = heat </span>
<span>m = mass of substance = 35.0 grams </span>
<span>C = 0.385 J/g*C </span>
<span>∆T = change in temperature = 65C - 20C = 45C </span>

<span>q = (35.0 g)*(0.385 J/g*C)*(45C) = 606 J </span>
3 0
3 years ago
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