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3 years ago

A) What mass in grams of H20 is needed to react completely with 40.0 g of

Chemistry

1 answer:

Step2247 [10]3 years ago

7 0

Answer:

$m_{H_2O}=9.24gH_2O$

Explanation:

Hello there!

In this case, since there is a 2:2 mole ratio between sodium peroxide and water according to the given reaction, it is possible to apply the following stoichiometric setup for the calculation of the required mass of water:

$m_{H_2O}=40.0gNa_O_2*\frac{1molNa_O_2}{78gNa_O_2}*\frac{2molH_2O}{2molNa_O_2} *\frac{18.02gH_2O}{1molH_2O} \\\\m_{H_2O}=9.24gH_2O$

Best regards!

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How many kilometers are there in 2,170.5 decimeters ?

Angelina_Jolie [31]

Answer:

0.21705

Explanation:

2170.5 decimeter=0.21705kilo

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3 years ago

What mass in grams (theoretical yield) of hydrogen is produced by the reaction of 31.3 g of

Verizon [17]

The percent yield is 74.4 %.

Fe₃O₄ + 4H₂ → 3Fe + 4H₂O

First, calculate the theoretical yield of H₂O.

Theor. yield =

0.112 g H

1 mol H

2.016 g H

4 mol H

18.02 g H

1 mol H

1.001 g H

Now calculate the percent yield.

% yield =

actual yield

theoretical yield

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3 years ago

Application - Equations and reacting masses

Nostrana [21]

1) 2 KCl + F₂ → 2 KF + Cl₂

2) 5.9 moles of F₂

3) 684.4 g of KF

4) 190 g of F₂

5) 0.526 g of KCl

6) 30.508 g of KF

7) 18.673 g of Cl₂

8) 50.362 g of KCl and 12.844 g of F₂

Explanation:

1) To balance the equation the number and type of moles entering the reaction have to be equal to the number and type of moles leaving the reaction.

2 KCl + F₂ → 2 KF + Cl₂

2) Knowing the chemical reaction we devise the following reasoning:

if 1 mole of F₂ produces 2 moles of KF

then X moles of F₂ produces 11.8 moles of KF

X = (1 × 11.8) / 2 = 5.9 moles of F₂

3) mass = number of moles × molecular weight

mass of KF = 11.8 × 58 = 684.4 g

4) mass = number of moles × molecular weight

mass of F₂ = 5 × 38 = 190 g

5) number of moles = mass / molecular weight

number of moles of F₂ = 10 / 38 = 0.263 moles

Knowing the chemical reaction we devise the following reasoning:

if 1 mole of F₂ react with 2 moles of KCl

then 0.263 moles of F₂ produces Y moles of KCl

Y = (0.263 × 2) / 1 = 0.526 g of KCl

6) Knowing the chemical reaction we devise the following reasoning:

if 1 mole of F₂ produces 2 moles of KF

then 0.263 moles of F₂ produces Z moles of KF

Z = (0.263 × 2) / 1 = 0.526 moles of KF

mass = number of moles × molecular weight

mass of KF = 0.526 × 58 = 30.508 g

7) Knowing the chemical reaction we devise the following reasoning:

if 1 mole of F₂ produces 1 mole of Cl₂

then 0.263 moles of F₂ produces W moles of Cl₂

W = (0.263 × 1) / 1 = 0.263 moles of Cl₂

mass = number of moles × molecular weight

mass of Cl₂ = 0.263 × 71 = 18.673 g

8) number of moles = mass / molecular weight

number of moles of KF = 39.2 / 58 = 0.676 moles

Knowing the chemical reaction we devise the following reasoning:

if 2 moles of KF are produced from 2 moles of KCl and 1 mole of F₂

then 0.676 moles of KF are produced from A moles of KF and B moles of F₂

A = (0.676 × 2) / 2 = 0.676 moles of KCl

B = (0.676 × 1) / 2 = 0.338 moles of F₂

mass = number of moles × molecular weight

mass of KCl = 0.676 × 74.5 = 50.362 g

mass of F₂ = 0.338 × 38 = 12.844 g

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moles

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3 years ago

Which process requires water to gain heat energy from th

Andrew [12]

Answer:

Evaporation, please mark me Brainliest

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3 years ago

Identify the group number in the periodic table of X, a representative element, in each of the following ionic compounds:

amm1812

In the formation of ionic compound, atoms of different elements that have obtained a nonneutral electrical charge (either due to losing or gaining electrons, creating a charge imbalance between the negatively-charged electrons and the positively-charged protons) are attracted in such a way that their interaction would result in electrical neutrality. Group 4A elements (the "carbon group") are a sort of dividing line: Elements in groups to the left of group 4A (the metallic elements) tend to have atoms that lose electrons during ionization, thus becoming more positive; elements in groups to the right of group 4A (the nonmetallic elements) tend to have atoms that gain electrons during ionization, thus becoming more negative. This is, of course, a broad generalization, but it should suffice in helping to get a big-picture idea of how to approach binary ionic compounds such as Al2X3.

Al, or aluminum, is a metallic element in group 3A, which is to the left of group 4A. A neutral aluminum atom has three valence electrons in its outermost shell. When an aluminum atom loses these three valence electrons (thus gaining a stable, noble gas electron configuration), it becomes a positively-charged aluminum ion, Al³⁺.

Now, let's consider how this aluminum cation (a positively-charged ion) could interact with anions (negatively-charged ions) to form an electrically-neutral compound. The most straightforward interaction would be an ionic bond between Al³⁺ and X³⁻. Why? Because the charges between the one Al³⁺ cation and the one X³⁻ anion are opposite and equal in magnitude; they would thus bond in a 1:1 unit ratio to form an ionic compound with the formula AlX.

But that's not what we have here! In our ionic compound, the Al has a subscript of 2, and the X has a subscript of 3. Here's the trick with ionic compounds: Assuming that our subscripts are all simplified to the lowest whole number, an ion's charge is equivalent in magnitude to the counterion's subscript (the counterion is simply the oppositely-charged ion in a binary ionic compound). The sign of the charge will be positive for the metal ion (which conventionally comes first in the formula unit) and negative for the nonmetal ion. The subscript of X is 3 in Al₂X₃, so the charge number on the Al ion is 3. Al, a metal, forms a positively-charged ion; thus, the Al ion has a 3+ charge.

Let's do the same for X. Al has a subscript of 2, so when the X atom ionizes, its charge must have a number of 2 and be negative, i.e., a 2- charge. Conveniently, the charge on an ion of a main-group and non-transition metal element tells us which group that ion's element is in. That's because the group number of elements in the s- and p-blocks is equal to the number of valence electrons in a neutral atom in that group. A group 1A element's atom has one valence electron; a group 2A element's atom has two valence electrons; and a group 3A element's atom (like an Al atom) has three valence electrons; and so forth.

An ion gains or loses electrons so that its outermost (valence) shell is completely filled. If element X's ion exists in the X²⁻ state, then that means it takes the addition of two electrons to give the X atom a full octet (eight electrons in the valence shell). We can then conclude that a neutral atom of X must have six valence electrons since 8 - 2 = 6. Thus, X is an element in group 6A (or group 16 on the extended periodic table), the chalcogens or "oxygen family."

Indeed, the most common form of the compound aluminum oxide has the formula Al₂O₃. Aluminum also forms an ionic compound with sulfur—another group 6A element—which has the formula Al₂S₃.

---

For XCO3, we must keep in mind that CO₃ as a whole is the counterion to X. The 3 is not the subscript for the CO₃ ion. CO₃, or carbonate, has a 2- charge: it can be written as (CO₃)²⁻. Here, the CO₃ ion as a group has no subscript, which might lead one to think that X must be an element from group 1A. Not so fast! It's critical to know beforehand that the carbonate ion has a 2- charge; if X were indeed a group 1A element, X would have a subscript of 2. But X has no subscript (a subscript of 1). Since the subscripts of ionic compounds are simplified to their lowest whole number ratios, that means that X must have a subscript of 2. Only then would both the subscripts of X and CO₃ be simplified to 1, as we have here. If CO₃ has a subscript of 2, that means that X is an element in group 2A (or group 2), the alkaline earth metals.

Indeed, the group 2A elemental ions magnesium (Mg²⁺) and calcium (Ca²⁺) both ionically bond with the carbonate ion to form magnesium carbonate and calcium carbonate, respectively; their respective formulae are MgCO3 and CaCO3, the same form as XCO3.

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3 years ago