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2 years ago

How many moles of hydrogen are in 5.2 moles of c7h18

2 answers:

alexandr1967 [171]2 years ago

6 0

The percentage of hydrogen in C7H18 is calculated as follows:
[18/(12*7+1*8)]*100=18%
The amount of hydrogen in 5.2moles is given by:(18/100)*5.2=0.94moles

alex41 [277]2 years ago

3 0

Answer:

There are 93.6 moles of hydrogen atoms are present in 5.2 moles of $C_7H_{18}$ .

Explanation:

We are given:

Moles of $C_7H_{18}$ = 5.2 mol

By Stoichiometry:

in 1 mole of $C_7H_{18}$ there are 18 moles of hydrogen atom.

So, moles of hydrogen atom in 5.2 mol of $C_7H_{18}$ :

$5.2\times 18 mol=93.6 mol$

There are 93.6 moles of hydrogen atoms are present in 5.2 moles of $C_7H_{18}$ .

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A uranium atom has 92 protons in it nucleus use what you know about atoms to predict how many electrons a neutral uranium atom h

sladkih [1.3K]

A neutral atom is one that has no net charge. In other words, the number of positive charges cancels out with the number of negative charges, so the number of protons equals the number of electrons. Uranium has 92 protons, which means it has a total positive charge of +92. To have a neutral charge, the atom MUST have 92 electrons, as that is a negative charge of -92, and the net charge is ultimately (protons + electrons) = ([+92] + [-92]) = 0.

6 0

3 years ago

In the laboratory a student combines 47.8 mL of a 0.321 M aluminum nitrate solution with 21.8 mL of a 0.366 M aluminum iodide so

alex41 [277]

Answer: The final concentration of aluminum cation is 0.335 M.

Explanation:

Given: $V_{1}$ = 47.8 mL (1 mL = 0.001 L) = 0.0478 L

$M_{1}$ = 0.321 M, $V_{2}$ = 21.8 mL = 0.0218 L, $M_{2}$ = 0.366 M

As concentration of a substance is the moles of solute divided by volume of solution.

Hence, concentration of aluminum cation is calculated as follows.

$[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}$

Substitute the values into above formula as follows.

$[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}\\= \frac{0.321 M \times 0.0478 L + 0.366 M \times 0.0218 L}{0.0478 L + 0.0218 L}\\= \frac{0.0153438 + 0.0079788}{0.0696}\\= 0.335 M$

Thus, we can conclude that the final concentration of aluminum cation is 0.335 M.

4 0

3 years ago

In a 5.609g of H20, how many molecules H20<br> arc<br> present?

nignag [31]

Answer:

1.88 x 10²³molecules of H₂O

Explanation:

Given parameters:

Mass of H₂O = 5.609g

Unknown:

Number of molecules of H₂O = ?

Solution:

To solve this problem, we need to find the number of moles of given compound first.

Number of moles = $\frac{mass}{molar mass}$

Molar mass of H₂O = 2(1) + 16 = 18g/mol

Number of moles = $\frac{5.609}{18}$ = 0.31mole

From mole concept;

1 mole of a substance contains 6.02 x 10²³molecules

0.31 mole of H₂O will therefore contain 0.31 x 6.02 x 10²³molecules

= 1.88 x 10²³molecules of H₂O

8 0

2 years ago

Una piedra es lanzada horizontalmente desde la azote de un edificio de 30 metros de altura, con una velocidad de 12 m/s. Hallar

drek231 [11]

Answer: La piedra tarda 2.47 segundos en caer al suelo, y cae a 29.64 metros de la base del edificio.

Explanation:

Ok, tenemos 2 sistemas de ecuaciones para este problema.

Primero, el vertical:

La aceleración es la aceleración gravitatoria, entonces:

a = -9.8m/s^2

para la velocidad podemos integrar sobre el tiempo y obtenemos

v = (-9.8m/s^2)*t + v0

donde v0 es la velocidad inicial, pero la velocidad inicial es solo horizontal, entonces v0 = 0.

Para la posición integramos de vuelta:

p = (1/2)*(-9.8m/s^2)*t^2 + p0

donde p0 es la posición inicial, en este caso, 30m

p = (-4.9m/s^2)*t^2 + 30m

la piedra va a llegar al piso cuando la posición vertical sea igual a 0m

p = 0m = (-4.9m/s^2)*t^2 + 30m

de aca podemos despejar el tiempo que la piedra tarda en llegar al suelo.

t = √(30/4.9) segundos = 2.47 s.

Ahora, las ecuaciones para el movimiento horizontal son:

Aceleración nula, pues no hay ninguna fuerza actuando en la dirección horizontal.

a = 0

para la velocidad, integramos sobre el tiempo:

v = 0*t + v0 = v0

donde v0 es la velocidad inicial, en este caso, es 12m/s

v = 12m/s

para la posición integramos de vuelta:

p = 12m/s*t + p0

en este caso podemos asumir que estamos inicialmente en el punto x = x0, asi que la posición inicial es p0 = x0.

p = 12m/s*t + x0

entonces, si queremos calcular la distancia entre la base del edificio y el punto donde cae la piedra, tenemos que calcular:

D = p(2.47s) - p(0s)

D = 12m/s*2.47s + x0 - (12m/s*0s + x0)

D = 29.64m

La piedra tarda 2.47 segundos en caer al suelo, y cae a 29.64 metros de la base del edificio.

3 0

2 years ago

An oxide of nitrogen is 25.9% N by mass, has a molar mass of 108 g/mol, and contains no nitrogen-nitrogen or oxygen-oxygen bonds

Sonja [21]

Answer:

The compound is N2O4

Explanation:

We have certain important pieces of information about the compound;

1) it is an oxide (a binary compound of nitrogen and oxygen)

2) there are no N-N bonds present

3) there are no O-O bonds present

Since it contains only nitrogen and oxygen then nitrogen accounts for 25.9% of the molecule by mass then oxygen should account for (100-25.9) = 74.1% oxygen

Relative atomic mass of oxygen = 16

Relative atomic mass of nitrogen = 14

We now deduce the empirical formula

Nitrogen. Oxygen

25.9/14. 74.1/16

1.85/1.85. 4.6/1.85 (divide through by the lowest ratio)

1 2

Empirical formula is NO2

To find the molecular formula

(NO2)n = 108

(14+2(16))n= 108

46n=108

n= 108/46

n= 2

Therefore molecular formula= N2O4

4 0

3 years ago