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How many grams of AuCl3, if the compound has 6.3 x10^23 atoms of Cl?

105.86 grams of AuCl3, if the compound has 6.3 x10^23 atoms of Cl.

Step-by-step explanation:

We are given that the compound has 6.3 x10^23 atoms of Cl.

To find how many molecules of AuCl3 are in the given compound, we divide the compound by 3, i.e;

 =   molecules of AuCl3.

Now, as we know that 1 mole of AuCI3 has molecules.

So, the moles that our compound has is given by;

      =    =  = 0.349 mole AuCI3

Also, the molar mass of AuCI3 = 303.33 g/mole

So, the molar mass of 0.349 moles AuCI3 =  

                                                                      =  105.86 g

Hence, 105.86 grams of AuCl3, if the compound has 6.3 x10^23 atoms of Cl.

93.8

step-by-step explanation:

lets say, first number = x

second number = 3x+7

total = 49

(x)+(3x+7)= 49

4x+7 = 49

4x = 42

x = 42/4

x= 10.5

x= 10.5

3x+7 = 3(10.5)+7 = 93.8

hope this

x> 6

step-by-step explanation:

8x+8> 7x+14

8x-7x> 14-8

x> 6