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umka2103 [35]
2 years ago
7

What is the molar mass for Agno3​

Chemistry
1 answer:
Aloiza [94]2 years ago
5 0

Answer:

169.87 g/mol

Explanation:

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I What<br>What is ment by<br>ment by covalency?​
Illusion [34]
When an element shares electrons with other atoms of the same or different elements to acquire stable electronic configuration, it is called covalency. If an atom shares 1 electron, its covalency is equal to 1. ... It needs 3 electrons to complete its octet.
5 0
2 years ago
At STP, how many moles of carbon dioxide will occupy 35.2 L of space?
Mrrafil [7]

Answer:

Moles of carbon dioxide are 1.57

Explanation:

Using ideal gas law, you can find moles of a gas with its pressure, temperature and volume, thus:

PV = nRT

PV/RT = n

<em>Where P is pressure, V is volume (35.2L); R is gas constant (0.082atmL/molK); T is absolute temperature and n are moles</em>

At STP, temperature is 273.15K and pressure is 1atm

Replacing:

1atm×35.2L / 0.082atmL/molK×273.15K = n

<em>1.57 = n</em>

<em>Moles of carbon dioxide are 1.57</em>

5 0
2 years ago
Match each of the following descriptions to thecorrect element listed
Slav-nsk [51]

Answer: Please see explanation for answer

Explanation:Matching the  descriptions to the correct element listed gives

Element Group Name Group No Metal/ Nonmetal/ Metalloid Description

Neon    Noble gases       18           Non metal Colorless, tasteless, and        odorless gas, unreactive, nonconductive.

Bromine      Halogens   17          Non metal Not found as a free element (uncombined) in nature, reddish-brown liquid that vaporizes readily at room temperature to a red gas with a strong disagreeable odor.

Beryllium       Alklaline earth metal  2    Metal  Silvery-white, ductile, malleable, conductive solid with a high melting point for this type of element

Platinum     Platimum group metals  10   Metal      Silvery white, relatively soft, low density, conductive solid that is not found as a free element in nature but commonly found combined in alloys with copper or nickel

Potassium   Alkali Metals   1   Metal  Soft, easily cut with a knife to expose a silvery surface that rapidly oxidizes in air; never found uncombined in nature

Silicon    Carbon family     1`4  Metalloid   Metallic luster and grayish solid, very common in rocks and gemstones such as amethyst and opal, semiconductor

Sulfur   Calcogens      16  Non metal  Pale yellow, odorless, brittle solid at room temperature, nonconductor

4 0
2 years ago
Using the equations
Anna [14]

Considering the Hess's Law, the enthalpy change for the reaction is 221.8 kJ/mol.

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

In this case you want to calculate the enthalpy change of:

C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g)

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: H₂ (g) + F₂ (g) → 2 HF (g)     ∆H° = -79.2 kJ/mol

Equation 2: C (s) + 2 F₂ (g) → CF₄ (g)     ∆H° = 141.3 kJ/mol

Equation 3: 2 C(s) + 2 H₂ (g) → C₂H₄ (g)     ∆H° = -97.6 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

<h3 /><h3>FIRST STEP</h3>

First, to obtain the enthalpy of the desired chemical reaction you need one mole of C₂H₄ (g) on reactant side and it is present in first equation. Since this equation has one mole of C₂H₄ (g) on the product side, it is necessary to locate it on the reactant side (invert it).

When an equation is inverted, the sign of ΔH° also changes.

<h3>SECOND STEP</h3>

Now, you need 2 moles of CF₄ (g) on the product side. The second equation has 1 mole of CF₄ (g) on the product side, so it is necessary to multiply it by 2 to obtain 2 moles of CF₄ (g).

Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 2, the variation of enthalpy also.

<h3>THIRD STEP</h3>

Finally, you need 4 moles of  HF (g) on the product side. The first equation has 2 moles of  HF (g) on the product side, so it is necessary to multiply it by 2 to obtain 4 moles of the compound.

Since the equation is multiply by 2, the variation of enthalpy also is multiplied by 2.

<h3>SUMMARY</h3>

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 H₂ (g) + 2 F₂ (g) → 4 HF (g)     ∆H° = -158.4 kJ/mol

Equation 2: 2 C (s) + 4 F₂ (g) → 2 CF₄ (g)     ∆H° = 282.6 kJ/mol

Equation 3: C₂H₄ (g) → 2 C(s) + 2 H₂ (g)     ∆H° = 97.6 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g)     ΔH°= 221.8 kJ/mol

Finally, the enthalpy change for the reaction is 221.8 kJ/mol.

Learn more about molar enthalpy:

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7 0
2 years ago
The reform reaction between steam and gaseous methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and dihyd
kenny6666 [7]

Answer:

The answer is "= 0.078 \ kg \ H_2".

Explanation:

calculating the moles in CH_4 =\frac{PV}{RT}

                                                =\frac{(0.58 \ atm) \times (923 \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(232^{\circ} C +273)}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(505)K}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (41.4605 \frac{L \cdot atm}{mol})}\\\\= 12.9 \ mol

Eqution:

CH_4 +H_2O \to  3H_2+ CO \ (g)

Calculating the amount of H_2 produced:

= 12.9 \ mol CH_4 \times  \frac{3 \ mol \ H_2 }{1 \ mol \ CH_4}\times \frac{2.016 g H_2}{1 \ mol \ H_2}\\\\= 78 \ g \ H_2 \\\\= 0.078 \ kg \ H_2

So, the amount of dihydrogen produced = 0.078 \frac{kg}{s}

5 0
3 years ago
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