Which orbital has the highest energy level

Which represents a linear graph?

Tpy6a [65]

Your answer is D. Hope this helps.
Linear graphs have representation as a STRAIGHT LINES.

7 0

3 years ago

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If the volume of 0.100M HCl delivered to 10.0mL of NaOH is 12.0mL the NaOH concentration is

iVinArrow [24]

<span>During neutralization reaction Milliequivalent of acid=milliequivalent of base. Milliequivalent of HCl=0.100*12*1 Milliequivalent of NaOH=M*10*1 So 0.100*12*1=M*10*1 Solve for M.</span>

3 0

3 years ago

How old is a bone in which the Carbon-14 in it has undergone 3 half-lives?

BartSMP [9]

5,700

That’s half

Hope that helps

5 0

3 years ago

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Assume that 50.0mL 50.0mL of 1.0MNaCl(aq) 1.0MNaCl(aq) and 50.0mL 50.0mL of 1.0M AgNO 3 (aq) 1.0MAgNO3(aq) were combined. Accord

S_A_V [24]

Answer:

The amount of precipitate formed would 7.175 grams of silver chloride.

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of NaCl = n

Volume of NaCl solution = 50.0 mL = 0.050 L

Molarity of the hydrogen peroxide = 2.0 M

$n=2.0 M\times 0.050 L=0.100 mol$

Moles of silver nitarte = n'

Volume of silver nitrate solution = 50.0 mL = 0.050 L

Molarity of the silver nitrate = 1.0 M

$n'=1.0 M\times 0.050 L=0.050 mol$

$NaCl(aq)+AgNO_3(aq)\rightarrow AgCl(s)+NaNO_3(aq)$

According to reaction, 1 mole of of silver nitrate reacts with 1 mole of NaCl. Then 0.050 mole of silve nitrate will :

$\frac{1}{1}\times 0.050 mol=0.050 mol$ of NaCl

This means that silver nitrate is in limiting amount and NaCl is in excessive amount.

So, the amount of AgCl depends upon amount of silver nitrate.

According to reaction, 1 mole of silver nitrate gives 1 mole of AgCl.

Then 0.050 moles of silver nitrate will give;

$\frac{1}{1}\times 0.050 mol=0.050 mol$ of AgCl

Mass of 0.050 moles of AgCl ;

$0.050 mol\times 143.5 g/mol=7.175 g$

The amount of precipitate formed would 7.175 grams of silver chloride.

8 0

3 years ago

You have 4 moles of oxygen gas in a flask. 4 moles of helium gas is added. What happens to the total pressure of the gases in th

ASHA 777 [7]

Answer: The correct option is (c). The total pressure doubles.

Solution:

Initially, only 4 moles of oxygen gas were present in the flask.

$p_{O_2}=Tp_1\times X_{O_2}$ ( $X_{O_2}=\frac{4}{4}$ ) ( according to Dalton's law of partial pressure)

$p_{O_2}=Tp_1\times 1=Tp_1$ ....(1)

$Tp_1$ = Total pressure when only oxygen gas was present.

Final total pressure when 4 moles of helium gas were added:

$X'_{O_2}=\frac{4}{8}=\farc{1}{2},X_{He}=\frac{4}{8}=\frac{1}{2}$

partial pressure of oxygen in the mixture :

Since, the number of moles of oxygen remains the same, the partial pressure of oxygen will also remain the same in the mixture.

$p_{O_2}=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}$

$Tp_2$ = Total pressure of the mixture.

from (1)

$Tp_1=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}$

On rearranging, we get:

$Tp_2=2\times Tp_1$

The new total pressure will be twice of initial total pressure.

7 0

3 years ago