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babunello [35]
2 years ago
14

What products can be formed from P + H2O + I2

Chemistry
1 answer:
torisob [31]2 years ago
7 0

The products that can be formed are hydrogen iodide and phosphoric acid.

<h3>What are reaction products?</h3>

They are substances formed from chemical reactions.

The equation of the illustrated reaction together with the products is as follows:

P + H_2O + I_2 --- > H_3PO_4 + HI

The products formed are phosphoric acid (H_3PO_4) and hydrogen iodide (HI.

Alternatively, the equation of the reaction can be written as:

18P + 5I_2 + 32H_2O -- > 10PH_4I + 8H_3PO_4

More on chemical reactions can be found here: brainly.com/question/22817140

#SPJ1

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We make a basic solution by mixing 50. mL of 0.10 M NaOH and 50. mL of 0.10 M Ca(OH)2. It requires 250 mL of an HCl solution to
andreyandreev [35.5K]

<u>Answer:</u> The correct answer is Option 5.

<u>Explanation:</u>

  • To calculate the molarity of the solution after mixing 2 solutions, we use the equation:

M=\frac{n_1M_1V_1+n_2M_2V_2}{V_1+V_2}

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of the NaOH.

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of the Ca(OH)_2

We are given:

n_1=1\\M_1=0.10M\\V_1=50mL\\n_2=2\\M_2=0.1\\V_2=50mL  

Putting all the values in above equation, we get:

M=\frac{(1\times 0.1\times 50)+(2\times 0.1\times 50)}{50+50}\\\\M=0.15M

  • To calculate the molarity of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base.

We are given:

n_1=1\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.15M\\V_2=100mL

Putting values in above equation, we get:

1\times M_1\times 250=1\times 0.15\times 100\\\\M_1=0.06M

Hence, the correct answer is Option 5.

7 0
3 years ago
If you found carbon 13 atom, you would know that
bearhunter [10]
The Options are as follow,

<span>a. it has 13 protons                  c. it has 13 neutrons                      </span>

<span>b. it has 13 electrons               d. it has 7 neutrons</span>


Answer:

             Option-d is the correct answer.


Explanation:

             As we know that C-13 has a atomic mass of 13. Also, atomic mass is given as,


<span>                 Atomic Mass  =  # of Protons  +  # of Neutrons   --- (1)</span>


Since;

          Atomic Mass  =  13

And,

          # of protons  =  6 (as number of protons  =  atomic number)


Putting values in equation 1,


                               13  =  6  -  # of Neutrons


Solving for Number of Neutrons,


                               # of Neutrons  =  13  -  6


                               # of Neutrons  =  7

3 0
2 years ago
Si hay luna nueva a la media noche del 1 de octubre, ¿cuando ocurrirá la próxima luna nueva?​
Nikitich [7]

Answer:

En el recién estrenado mes de octubre comienza a notarse cómo los días se acortan y las noches se extienden. El efecto se hará más plausible todavía cuando entre el horario de invierno el próximo domingo 25. Ese día los relojes se atrasarán una hora y a las 3:00 serán las 2:00.

Las cada vez más dilatadas noches aumentan el protagonismo de la Luna. El satélite natural de la Tierra comenzará y acabará el mes por todo lo alto con dos brillantes lunas llenas. El mes lunar dura 29,5 días y los meses solares duran por lo menos 30, salvo febrero. Así que echen cuentas: en alguno tendrían que coincidir dos fases llenas. Por cierto, la noche del 31 es la noche de Halloween. Los disfraces de licántropo tendrán el complemento perfecto.

Según el Observatorio Astronómico Nacional, el día 1 la luna llena será a las 23:05 y el 31 a las 15:49. Que se fije una hora concreta es similar a lo que ocurre con los equinoccios: la Luna está en movimiento, no se para. Así que a partir de ese punto, el satélite comienza su siguiente fase. La luna nueva tendrá lugar el 16 de octubre a las 21:31.

A veces se observa el satélite durante el día. Si es así, significa que está en fase creciente o menguante. Este mes el cuarto menguante ocurrirá el día 10 a las 2:40. El 23 de octubre a las 15:23 será el turno del creciente.

Tanto de día como de noche, la Luna puede observarse a simple vista y sin dificultad alguna. Con unos buenos prismáticos se llegan a ver además algunos detalles de su superficie, como cráteres o “mares”. Más difícil de observar son otros fenómenos que también tendrán lugar en octubre, por ejemplo, las lluvias de estrellas. No resultarán tan llamativas como las famosas Perseidas estivales, pero siempre se puede cazar algún meteoro con la vista.

Las más intensas serán las Oriónidas del 21 de octubre, provenientes de restos del cometa Halley. Hasta 20 meteoros cruzando el cielo por hora en esta fecha central. En las noches anteriores y posteriores también podrá verse alguna estrella fugaz. Antes de la Oriónidas, tendrán lugar las Dracónidas, aunque menos vistosas. Unos 10 meteoros por hora serán los que se aprecien en la jornada central del 8 de octubre

8 0
2 years ago
For the following reaction, 53.7 grams of iron(III) oxide are allowed to react with 22.8 grams of aluminum. iron(III) oxide (s)
ZanzabumX [31]

Answer:

34.23 grams is the maximum amount of aluminum oxide that can be formed.

Iron (III) oxide is a limiting reagent i.e Fe_2O_3.

4.6764 grams is the amount of the aluminum which remains after the reaction is complete

Explanation:

iron(III) oxide (s) + aluminum (s) → aluminum oxide (s) + iron (s)

Fe2O_3(s)+2Al(s)\rightarrow Al_2O_3(s)+2Fe(s)

Moles of  iron(III) oxide : \frac{53.7 g}{160 g/mol}=0.3356 mol

Moles of aluminium : \frac{22.8 g}{27 g/mol}=0.8444 mol

According to recation, 1 mole of iron(III) oxide reacts with 2 moles of aluminum.

Then 0.3356 moles of iron(III) oxide will react with:

\frac{2}{1}\times 0.3356 mol=0.6712 mol of aluminum.

As we can see that moles of iron(III) are in limiting amount.Hence iron(III) oxide is a limiting reagent i.e Fe_2O_3 and aluminum in the an excessive reagent.

Amount of aluminum oxide will depend upon moles of limiting reagent that is iron(III) oxide.

According to reaction , 1 mole iron(III) oxide gives 1 moles of aluminum oxide.

Then 0.3356 moles will give:

\frac{1}{1}\times 0.3356 mol=0.3356 mol of aluminum oxide

Mass of 0.3356 moles of aluminum oxide:

0.3356 mol × 102 g/mol = 34.23 g

34.23 grams is the maximum amount of aluminum oxide that can be formed.

Moles of excessive reagent left = 0.8444 mol - 0.6712 mol = 0.1732 mol

Mass of 0.1732 moles of aluminum :

0.1732 mol × 27 g/mol = 4.6764 g

4.6764 grams is the amount of the aluminum which remains after the reaction is complete

7 0
3 years ago
Is nitrogen reactive or stable and why???
kolbaska11 [484]
I think it would be pretty reactive, because it is flammable but it need something  to set it off like fire or small explosion.
4 0
3 years ago
Read 2 more answers
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