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Talja [164]
2 years ago
7

What mass of hydrogen gas can be produced by 6.0 mol of aluminum with hydrochloric acid?

Chemistry
1 answer:
Agata [3.3K]2 years ago
3 0
2Al + 6HCl = 2AlCl₃ + 3H₂

m(H₂)=3M(H₂)n(Al)/2

m(H₂)=3*2.0g/mol*6.0mol/2=18.0 g
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Nikolas, the fire extinguisher, and the skateboard have a combined mass of 50 kg. What force would the fire extinguisher have to
murzikaleks [220]

Answer:

Force used by fire extinguisher = 60 N

Explanation:

Given:

Mass of skateboard with fire extinguisher = 50 kg

Acceleration of fire extinguisher = 1.2 m/s²

Find:

Force used by fire extinguisher = ?

Computation:

⇒ Force = Mass × Acceleration

⇒ Force used by fire extinguisher = Mass of skateboard with fire extinguisher × Acceleration of fire extinguisher

⇒ Force used by fire extinguisher = 50 kg × 1.2 m/s²

⇒ Force used by fire extinguisher = 60 N

3 0
3 years ago
Read 2 more answers
The by-product of the chlorination of an alkane is​
Elanso [62]

The by-product of the chlorination of an alkane is​  <u>HCl</u>

Explanation:

  • Chlorination is the process of adding chlorine to drinking water to disinfect it and kill germs. Different processes can be used to achieve safe levels of chlorine in drinking water.
  • Chlorination of alkane gives a mixture of different products.
  • When consider mechanism of alkanes chlorination, free radicals are formed during the reaction to keep the continuous reaction.
  • Different alkyl chloride compounds, extended carbon chains compounds and HCl are formed as products in product mixture.
  • Chlorination byproducts, their toxicodynamics and removal from drinking water.
  • Halogenated trihalomethanes (THMs) and haloacetic acids (HAAs) are two major classes of disinfection byproducts (DBPs) commonly found in waters disinfected with chlorine
  • Chlorine is available as compressed elemental gas, sodium hypochlorite solution (NaOCl) or solid calcium hypochlorite (Ca(OCl)2
7 0
3 years ago
A sample of a compound contains 41.33 g of carbon and 8.67 g of hydrogen. The molar mass of the
mihalych1998 [28]

Answer:

C6H15

Explanation:

First, we calculate the empirical formula as follows:

C = 41.33 g

H = 8.67 g

We convert each mass value to mole by dividing each element by its molar mass (C = 12g/mol, H = 1g/mol)

C = 41.33g ÷ 12g/mol = 3.44mol

H = 8.67g ÷ 1g/mol = 8.67mol

Next, we divide each mole value by the smallest (3.44mol)

C = 3.44mol ÷ 3.44mol = 1

H = 8.67mol ÷ 3.44mol = 2.52

We multiply this ratio by 2 to get a simple whole number ratio

C = 1 × 2 = 2

H = 2.52 × 2 = 5.04

Based on this, the whole number ratio of C and H is 2:5, hence, the empirical formula is C2H5.

The molecular mass of the compound is given as 87.18 g/mol, hence, the molecular formula is calculated as follows:

(C2H5)n = 87.18

[12(2) + 1(5)]n = 87.18

[24 + 5]n = 87.18

(29)n = 87.18

n = 87.18 ÷ 29

n = 3.006

Approximately to whole number, n = 3

Hence, the molecular formula of the compound is [C2H5]3

= C6H15

4 0
2 years ago
The atomic mass of helium-4 is 4.0026 amu. How many of each
umka21 [38]

Answer:

Protons: 2.

Electrons: 2.

Neutrons: 2.

Explanation:

Hello,

In this case, since an atom's atomic number is equal to the number of electrons, considering the electron configurations, taking into account that helium-4 is neither positively nor negatively charged, we can infer that the number of electrons equal the number of protons, which in this case are 2, due to the fact that is atomic number is 2.

Moreover, as helium-4's atomic number is 4 as a whole number, we compute the number of neutrons by using the shown below equation:

Neutrons=mass\ number-atomic\ number\\\\Neutrons=4-2\\\\Neutrons=2

Regards.

7 0
2 years ago
A 360. g iron rod is placed into 750.0 g of water at 22.5°C. The water temperature rises to 46.7°C. What was the initial tempera
Grace [21]

Answer: The initial temperature of the iron was 515^0C

Explanation:

heat_{absorbed}=heat_{released}

As we know that,  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]         .................(1)

where,

q = heat absorbed or released

m_1 = mass of iron = 360 g

m_2 = mass of water = 750 g

T_{final} = final temperature = 46.7^0C

T_1 = temperature of iron = ?

T_2 = temperature of water = 22.5^oC

c_1 = specific heat of iron = 0.450J/g^0C

c_2 = specific heat of water= 4.184J/g^0C

Now put all the given values in equation (1), we get

-360\times 0.450\times (46.7-x)=[750\times 4.184\times (46.7-22.5)]

T_i=515^0C

Therefore, the initial temperature of the iron was 515^0C

4 0
3 years ago
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