A certain element forms an ion with 54 electrons and a charge of +2. identify the element.

zheka24 [161]

Answer:

2.14 × 10⁻³ molecules/RSP

3.31 × 10⁻³ molecules/ESP

Explanation:

Step 1: Calculate the number of moles of Acetaminophen per Regular Strength Pill (RSP)

A Regular Strength Pill has 1.29 × 10²¹ molecules of Acetaminophen per pill. To convert molecules to moles we will use Avogadro's number: there are 6.02 × 10²³ molecules in 1 mole of molecules.

1.29 × 10²¹ molecules/RSP × 1 mol/6.02 × 10²³ molecules = 2.14 × 10⁻³ molecules/RSP

Step 2: Calculate the number of moles of Acetaminophen per Extra Strength Pill (ESP)

An Extra Strength Pill has 1.99 × 10²¹ molecules of Acetaminophen per pill. To convert molecules to moles we will use Avogadro's number: there are 6.02 × 10²³ molecules in 1 mole of molecules.

1.99 × 10²¹ molecules/ESP × 1 mol/6.02 × 10²³ molecules = 3.31 × 10⁻³ molecules/ESP

5 0

2 years ago

How many moles of carbon atoms is 7.72 x 10 ^24 atoms of carbon? Round to 2 decimal places

soldier1979 [14.2K]

Answer:

<h2>12.82 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

$n = \frac{7.72 \times {10}^{24} }{6.02 \times {10}^{23} } \\ = 12.8 239$

We have the final answer as

<h3>12.82 moles</h3>

Hope this helps you

5 0

2 years ago

3. Methyl acetate is hydrolyzed at 25 oC in acidic environment. Aliquots of equal volume are removed and titrated with NaOH solu

avanturin [10]

Solution :

Time (sec) Volume of NaOH (mL)

339 26.23

1242 27.80

2745 29.70

4546 3.81

$\infty$ 39.81

Now the example of the first order kinetics w.r.t volumetric analysis is :

$k=\frac{2.303}{t} \log \left(\frac{v_{\infty}-v_0}{v_{\infty}- v_t}\right)$

Here, $v_{\infty}= \text{ volume at }\infty = 39.81$

$v_{t}= \text{ volume at time 't' } = 27.80$

$v_0$ = volume at time 0 = 0

Since the interval is not constant, we take the time interval as

$=\frac{903+1503+1801}{3}$

$=\frac{4207}{3}$

= 1402.3333

≈ 1402 seconds

$k=\frac{2.303}{1402} \log \left(\frac{39.81-0}{39.81-27.80}\right)$

$=(0.001643) \log \left(\frac{39.81}{10.01}\right)$

= 0.001643 x 0.52045

= 0.00082

$= 8.55 \times 10^{-4} \ sec^{-1}$

Therefore, the first order rate constant is k $= 8.55 \times 10^{-4} \ sec^{-1}$ .

5 0

2 years ago

Identify the products in the reaction HCl+NaOH=NaCl+H2O

valentina_108 [34]

NaCl and H2O.

The products are typically the elements/compounds on the right side of the equation or the right side of the arrow. The left side of the arrow would be the reactants of the equation.

Hope this helps!

5 0

2 years ago

When a solution of 0.1 M Mg(NO3)2 was mixed with a limited amount of aqueous ammonia, a light white, wispy solid was observed, i

Ipatiy [6.2K]

<u>Answer:</u> The net ionic equation is written below.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of magnesium nitrate and aqueous ammonia (ammonium hydroxide) is given as:

$Mg(NO_3)_2(aq.)+2NH_4OH(aq.)\rightarrow Mg(OH)_2(s)+2NH_4NO_3(aq.)$

A white precipitate of magnesium hydroxide is formed in the above reaction.

Ionic form of the above equation follows:

$Mg^{2+}(aq.)+2NO_3^-(aq.)+2NH_4^+(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)+2NH_4^+(aq.)+2NO_3^-(aq.)$

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Mg^{2+}(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)$

Hence, the net ionic equation is written above.

8 0

2 years ago