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Kaylis [27]
3 years ago
10

How many atoms of hydrogen are in 0.500 mol of ch3oh molecules?

Chemistry
1 answer:
cestrela7 [59]3 years ago
8 0
In 1 mol of CH3OH, you have 4 H-atoms (because 3 H-atoms are attached to the C-atom, and one H-atom in the OH group). That means in 0.500 mol of CH3OH, you have 2 H-atoms since it is halved. And then we have Avogadro's constant: 6.02 * 1023.

The question asks for how many hydrogen atoms there are in 0.500 mol CH3OH. Using the numbers that we have (Avogadro's constant and no. of H-atoms), the answer of the question will be something like:

<span>H-atoms in CH3OH = 2 * 6.02 * </span>1023<span> = ~1.2 * 10</span>24

 


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In the formula for the compound XCl4, the X could represent<br> (1) C (3) Mg<br> (2) Η (4) Zn
Anvisha [2.4K]
Correct Answer: option 1 i.e. C

Reason: 
The the compound of interest i.e.  XCl4, since there are 4 Cl atoms bonded to X. This signifies that the valency of X is 4.

There atomic number of C is 6. It's electronic configuration is giving by 1s2 2s2 2p2. Thus, there are 4 electrons in valence shell of C. This signifies that valency of C is 4. Hence the compound present in present case is CCl4.
7 0
4 years ago
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if 1.386 g of mg ribbon combusts to form 2.309 g of oxide product, calculate the experimental mass percent of oxygen from this d
Pavlova-9 [17]

1.386 g of Mg ribbon combusts to form 2.309 g of oxide product. The mass percent of oxygen in the oxide is 40.0 %.

Let's consider the reaction for the combustion of Mg.

Mg + 1/2 O₂ ⇒ MgO

1.386 g of Mg combusts to form 2.309 g of MgO. We want to determine the mass of oxygen in MgO. According to Lavoisier's law of conservation of mass, matter is not created nor destroyed over the course of a chemical reaction. Then, the mass of Mg in the reactants is equal to the mass of Mg in MgO. The mass of the magnesium oxide is the sum of the masses of magnesium and oxygen. The <u>mass of oxygen in the oxide</u> is:

mMgO = mMg + mO\\mO = mMgO - mMg = 2.309 g - 1.386 g = 0.923 g

We can calculate the mass percent of O in MgO using the following expression.

\% O = \frac{mO}{mMgO} \times 100\% = \frac{0.923 g}{2.309g} \times 100\%  = 40.0 \%

You can learn more about mass percent here: brainly.com/question/14990953

3 0
2 years ago
Which models of the atom in task 1 are not supported by the results of the hydrogen gas experiment? For each of these models, ex
kramer

Answer:

Thomson  placed two magnets on either side of the tube, and observed that this magnetic field also deflected the cathode ray. The results of these experiments helped Thomson determine the mass-to-charge ratio of the cathode ray particles, which led to a fascinating discovery, minus the mass of each particle was much, much smaller than that of any known atom. Thomson repeated his experiments using different metals as electrode materials, and found that the properties of the cathode ray remained constant no matter what cathode material they originated from. From this evidence, Thomson made the following conclusions:

The cathode ray is composed of negatively-charged particles.

The particles must exist as part of the atom, since the mass of each particle is only ~1/2000 the mass of a hydrogen atom.

These subatomic particles can be found within atoms of all elements.

While controversial at first, Thomson's discoveries were gradually accepted by scientists. Eventually, his cathode ray particles were given a more familiar name: electrons. The discovery of the electron disproved the part of Dalton's atomic theory that assumed atoms were indivisible. In order to account for the existence of the electrons, an entirely new atomic model was needed.

Explanation:

8 0
2 years ago
How many carbon atoms are there in a diamond (pure carbon) with a mass of 55 mg ?
kari74 [83]
1 mol of Carbon = 12 grams.
x mol of Carbon = 55 grams

12*x = 1 * 55
x = 55/12
x = 4.583333 mols of carbon

1 mol of anything is 6.02 * 10^23 atoms
4.58333333 mol = x

1/4.5833333 = 6.02 * 10^23/x
x = 4.58333* 6.02*10^23
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3 years ago
You to cute ✌✌<br><br> i will give u brailtist if u send me something funny.
Paul [167]

Answer:

Here some things

Explanation:

8 0
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