Question 2 (1 point)

34 atoms of carbon (C) react with 22 molecules of hydrogen gas (H2). How many molecules of methane (CH4) will be formed, and wha

kolbaska11 [484]

Answer:

11 molecules of CH4.

23 atoms of C is the leftover.

Explanation:

Hello!

In this case, for the formation of methane:

$C+2H_2\rightarrow CH_4$

We can see there is an excess of carbon based on their stoichiometry, because the needed amount of hydrogen gas molecules would be:

$molecules _{H_2}=34atomC*\frac{2molec\ H_2}{1atomC} =68molec\ H_2$

Thus, the formed molecules of methane are computed below:

$molec\ CH_4=22molec\ H_2 *\frac{1molec\ CH_4}{2molmolec\ H_2} \\\\molec\ CH_4=11molec\ CH_4$

In such a way, the leftover of carbon atoms are:

$atoms \ C^{left over}=34-22molec\ H_2*\frac{1atoms C}{2molec\ H_2} \\\\atoms \ C^{left over}=23 atoms C$

Best regards!

4 0

3 years ago

What volume will 12.0 g of oxygen gas occupy at 25 c and a pressure of 52.7 kpa?

Stolb23 [73]

We can use the ideal gas law equation to find the volume occupied by oxygen gas
PV = nRT
where ;
P - pressure - 52.7 kPa
V - volume
n - number of oxygen moles - 12.0 g / 32 g/mol = 0.375 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 25 °C + 273 = 298 K
substituting the values in the equation
52 700 Pa x V = 0.375 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
V = 17.6 L
volume of the gas is 17.6 L

8 0

3 years ago

Use the following graph of a car traveling on a straight northerly path to answer this question. At what time would the

BARSIC [14]

Answer:

B 144.0 s is the best answer of this question

6 0

4 years ago

27. The density of nickel is 8.91 g/cm3. How large a cube, in cm3, would contain 2.00 x 10^24 atoms of nickel? Use dimensional a

jok3333 [9.3K]

Answer : The volume of the cube is, $21.88cm^3$

Solution : Given,

Density of nickel = $8.91g/cm^3$

Number of nickel atoms = $2\times 10^{24}$

Molar mass of nickel = 58.7 g/mole

First we have to calculate the moles of nickel.

As, $6.022\times 10^{23}$ atoms form 1 mole of nickel

So, $2\times 10^{24}$ atoms form $\frac{2\times 10^{24}}{6.022\times 10^{23}}=3.321$ moles of nickel

The moles of nickel = 3.321 moles

Now we have to calculate the mass of nickel.

$\text{ Mass of Ni}=\text{ Moles of Ni}\times \text{ Molar mass of Ni}$

$\text{ Mass of Ni}=(3.321moles)\times (58.7g/mole)=194.94g$

The mass of nickel = 194.94 g

Now we have to calculate the volume of nickel.

$Density=\frac{Mass}{Volume}$

$8.91g/cm^3=\frac{194.94g}{Volume}$

$Volume=21.88cm^3$

Therefore, the volume of the cube is, $21.88cm^3$

4 0

3 years ago

How many orbitals surround the nucleus in a neutral atom of sulfur (S)?

Leto [7]

Explanation:

The atomic number of S is $16$

So,number of electrons in S is $16$

The electronic configuration of S is $2,8,6$

The orbital electronic configuration of S is $^{1}s_{2}^\text{ }^{2}s_{2}^\text{ }^{2}p_{x2}^\text{ }^{2}p_{y2}^\text{ }^{2}p_{z2}^\text{ }^{3}s_{2}^\text{ }^{3}p_{x2}^\text{ }^{3}p_{y1}^\text{ }^{3}p_{z1}^\text{ }$

So,the number of orbitals involved is 9.

3 0

3 years ago