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Gala2k [10]
3 years ago
11

PLS help!!!!!!!! Why are halogen atoms never a central atom?

Chemistry
2 answers:
Scilla [17]3 years ago
4 0

Answer:

They can’t have more than 8 valence electrons.

Explanation:

gayaneshka [121]3 years ago
3 0
They can’t have more than 8 valence electrons. Since fluorine is in group 7 it can only bound to one other molecule
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How many moles of dinitrogen monoxide are present in 9.3 x 10^22 molecules of this compound?
Step2247 [10]

Answer:

<h2>0.15 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities.

From the question we have

n =  \frac{9.3 \times  {10}^{22} }{6.02 \times  {10}^{23} }  \\  = 0.154485...

We have the final answer as

<h3>0.15 moles</h3>

Hope this helps you

8 0
3 years ago
Helpppp<br> pleasee!!!!! :)
frez [133]
Weight cost toxicity
4 0
3 years ago
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A mixture contains NaHCO3 together with unreactive components. A 1.75 g sample of the mixture reacts with HA to produce 0.561 g
Lynna [10]

Answer:

\%NaHCO_3=61.2\%

Explanation:

Hello.

In this case, since the undergoing chemical reaction is only between the sodium bicarbonate and the acid HA:

NaHCO_3+HA\rightarrow NaA+H_2O+CO_2

For 0.561 g of yielded carbon dioxide (molar mass 44 g/mol), the following mass of sodium bicarbonate (molar mass 84 g/mol) that reacted was:

m_{NaHCO_3}=0.561gCO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molNaHCO_3}{1molCO_2} *\frac{84gNaHCO_3}{1molNaHCO_3} \\\\m_{NaHCO_3}=1.071g

Considering the 1:1 mole ratio between CO2 and NaHCO3. Finally, the percent by mass of NaHCO3 is computed by dividing the mass of reacted NaHCO3 and t the mixture:

\%NaHCO_3=\frac{1.071g}{1.75g}*100\%\\ \\\%NaHCO_3=61.2\%

Best regards.

5 0
3 years ago
which of the following compounds will likely have a bent triatomic structure? A. H2S. B. O2 C.NI3 D.CH4
Ymorist [56]

a) H2S is your answer.

5 0
3 years ago
A patient gets 2.1 L of fluid over 19 hours through an IV. The drop factor is 20 gtt/mL.Calculate the drip rate in drops per min
Aleks [24]
<span>The rate of infusion is 2.1L/19h or 2100mL/19h (as 1L = 100 mL).
       
To convert 19 hours to minutes we multiply as follows:
   
19 hours = (19 hours) x (60 minutes/1 hour) = 1140 minutes
       
So the rate of infusion becomes:
   
2100mL /1140 min
       
In order to converted mL to drops (gtt) we multiply the rate of infusion with the drop factor to get the drip rate:
       
(2100mL/1140min) x (20 gtt/mL) = 36.8 gtt/min</span>
8 0
3 years ago
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