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2 years ago

Please help !!!

1 answer:

3 0

Answer: 12 neutrons
Explanation: The mass number of an element tells us the number of protons AND neutrons in an atom (the two particles that have a measurable mass). Sodium has a mass number of 23amu. Since sodium has 11 protons, the number of neutrons must be 23 – 11 = 12 neutrons.

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How many liters of ammonia (NH3), at 3.2 atm and 23C, must be used to produce of 2.65 grams of calcium hydride (CaH2). 6 Ca(s)

7nadin3 [17]

Answer:

The answer to your question is V = 0.32 L

Explanation:

Data

Volume of NH₃ = ?

P = 3.2 atm

T = 23°C

mass of CaH₂ = 2.65 g

Balanced chemical reaction

6Ca + 2NH₃ ⇒ 3CaH₂ + Ca₃N₂

Process

1.- Convert the mass of CaH₂ to moles

-Calculate the molar mass of CaH₂

CaH₂ = 40 + 2 = 42 g

42 g ------------------ 1 mol

2.65 g -------------- x

x = (2.65 x 1)/42

x = 0.063 moles

2.- Calculate the moles of NH₃

2 moles of NH₃ --------------- 3 moles of CaH₂

x --------------- 0.063 moles

x = (0.063 x 2) / 3

x = 0.042 moles of NH₃

3.- Convert the °C to °K

Temperature = 23°C + 273

= 296°K

4.- Calculate the volume of NH₃

-Use the ideal gas law

PV = nRT

-Solve for V

V = nRT / P

-Substitution

V = (0.042)(0.082)(296) / 3.2

-Simplification

V = 1.019 / 3.2

-Result

V = 0.32 L

7 0

2 years ago

How many grams of water are produced when 2.50 mol oxygen reacts with hydrogen?

natali 33 [55]

I think the answer is D

4 0

2 years ago

Calculate the pOH of an aqueous solution of .0.073 M LiOH

Novosadov [1.4K]

Considering the definition of pOH and strong base, the pOH of the aqueous solution is 1.14

The pOH (or potential OH) is a measure of the basicity or alkalinity of a solution and indicates the concentration of ion hydroxide (OH-).

pOH is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:

pOH= - log [OH⁻]

On the other hand, a strong base is that base that in an aqueous solution completely dissociates between the cation and OH-.

LiOH is a strong base, so the concentration of the hydroxide will be equal to the concentration of OH-. This is:

[LiOH]= [OH-]= 0.073 M

Replacing in the definition of pOH:

pOH= -log (0.073 M)

In summary, the pOH of the aqueous solution is 1.14

Learn more:

7 0

2 years ago

Which particle in an atom is lost or gained to make an ion

denpristay [2]

An electron because that is the only part able to be lost or gained without nuclear action needed

7 0

2 years ago

If 45.00 g of precipitate is formed from the reaction of 0.100 mol/L

Tom [10]

Answer:

Approximately $2.53\; \rm L$ (rounded to three significant figures) assuming that ${\rm HCl}\, (aq)$ is in excess.

Explanation:

When ${\rm HCl} \, (aq)$ and ${\rm AgNO_3}\, (aq)$ precipitate, ${\rm AgCl} \, (s)$ (the said precipitate) and $\rm HNO_3\, (aq)$ are produced:

${\rm HCl}\, (aq) + {\rm AgNO_3}\, (aq) \to {\rm AgCl}\, (s) + {\rm HNO_3}\, (aq)$ (verify that this equation is indeed balanced.)

Look up the relative atomic mass of $\rm Ag$ and $\rm Cl$ on a modern periodic table:

$\rm Ag$ : $107.868$ .
$\rm Cl$ : $35.45$ .

Calculate the formula mass of the precipitate, $\rm AgCl$ :

$\begin{aligned}& M({\rm AgCl})\\ &= (107.868 + 35.45)\; \rm g \cdot mol^{-1} \\\ &\approx 143.318 \; \rm g\cdot mol^{-1}\end{aligned}$ .

Calculate the number of moles of $\rm AgCl$ formula units in $45.00\; \rm g$ of this compound:

$\begin{aligned}n({\rm AgCl}) &= \frac{m({\rm AgCl})}{M({\rm AgCl})} \\ &\approx \frac{45.00\; \rm g}{143.318\; \rm g \cdot mol^{-1}}\approx 0.313987\; \rm mol \end{aligned}$ .

Notice that in the balanced equation for this reaction, the coefficients of ${\rm AgNO_3} \, (aq)$ and ${\rm AgCl}\, (s)$ are both one.

In other words, if ${\rm HCl}\, (aq)$ (the other reactant) is in excess, it would take exactly $1\; \rm mol$ of ${\rm AgNO_3} \, (aq)\!$ formula units to produce $1\; \rm mol \!$ of ${\rm AgCl}\, (s)\!$ formula units.

Hence, it would take $0.313987\; \rm mol$ of ${\rm AgNO_3} \, (aq)\!$ formula units to produce $0.313987\; \rm mol\!$ of ${\rm AgCl}\, (s)\!$ formula units.

Calculate the volume of the ${\rm AgNO_3} \, (aq)\!$ solution given that the concentration of the solution is $0.124\; \rm mol \cdot L^{-1}$ :

$\begin{aligned}V({\rm AgNO_3}) &= \frac{n({\rm AgNO_3})}{c({\rm AgNO_3})} \\ &\approx \frac{0.313987\; \rm mol}{0.124\; \rm mol \cdot L^{-1}}\approx 2.53\; \rm L\end{aligned}$ .

(The answer was rounded to three significant figures so as to match the number of significant figures in the concentration of ${\rm AgNO_3} \, (aq)\!$ .)

In other words, approximately $2.53\; \rm L$ of that ${\rm AgNO_3} \, (aq)\!$ solution would be required.

4 0

2 years ago