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Cerrena [4.2K]
3 years ago
14

Which gives up the electrons,Lithium or chlorine?

Chemistry
2 answers:
olya-2409 [2.1K]3 years ago
8 0
Lithium gives up electrons
coldgirl [10]3 years ago
5 0
Hey there,
Lithum<span> atoms readily </span>give up<span> one </span>electron to form positively charged. W<span>hile </span>chlorine<span> is bonded to sodium ionically in sodium chloride,
</span>Thus, Lithium gives up electrons.

Hope this helps :))

~Top
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Calculate the volume which 1.00 mole of a gas occupies at 1 atm and 298K?
elena-14-01-66 [18.8K]

Answer:

25.45 Liters

Explanation:

Using Ideal Gas Law PV = nRT => V = nRT/P

V = (1mole)(0.08206Latm/molK)(298K)/(1atm) = 25.45 Liters

7 0
2 years ago
How many grams of the excess reactant are left over according to the reaction below given that you start with 10.0 g of Al and 1
valentinak56 [21]
<span>4 Al + 3 O2 → 2 Al2O3 

(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al 
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2 

0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess. 

((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) = 
10.1 g O2 left over</span><span>
</span>
7 0
3 years ago
Read 2 more answers
Which factors influence the salinity of ocean surface water?
In-s [12.5K]
C and D

surface water is evaporated by the sun and concentrated the ocean salt and oceans are diluted when rivers flow into them
4 0
3 years ago
Read 2 more answers
Determine the empirical formula of the following compound if a sample contains 0.104 molK, 0.052 molC, and 0.156 molO;?
faltersainse [42]

Answer:

K₂CO₃    

Explanation:

Given parameters:

Number of moles of K = 0.104mol

Number of moles of C = 0.052mol

Number of moles of O = 0.156mol

Method

From the given parameters, to calculate the empirical formula of the elements K, C and O, we reduce the given moles to the simplest fraction.

Empirical formula is the simplest formula of a compound and it differs from the molecular formula which is the actual formula of a compound.

  • Divide the given moles through by the smallest which is C, 0.052mol.
  • Then approximate values obtained to the nearest whole number of multiply by a factor to give a whole number ratio.
  • This is the empirical formula

Solution

Elements                             K                       C                    O

Number of moles            0.104                0.052            0.156

Dividing by the

smallest                       0.104/0.052     0.052/0.052  0.156/0.052

                                            2                           1                     3

The empirical formula is K₂CO₃      

3 0
2 years ago
An unknown piece of metal weighing 95.0 g is heated to 98.0°C. It is dropped into 250.0 g of water at 23.0°C. When equilibrium i
lutik1710 [3]

Answer:

C_{metal}=126.6\frac{J}{g\°C}

Explanation:

Hello!

In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:

Q_{metal}=-Q_{water}

Which can be also written as:

m_{metal}C_{metal}(T_{EQ}-T_{metal})=-m_{water}C_{water}(T_{EQ}-T_{water})

Thus, since we need the specific heat of the metal, we solve for it as shown below:

C_{metal}=\frac{m_{water}C_{water}(T_{EQ}-T_{water})}{-m_{metal}(T_{EQ}-T_{metal})} \\\\C_{metal}=\frac{250.0g*4.184\frac{J}{g\°C}(29.0\°C-98.0\°C)}{95.0g(29.0\°C-23.0\°C)} \\\\C_{metal}=126.6\frac{J}{g\°C}

Best regards.

7 0
2 years ago
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