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2 years ago

I have 0.725 liters of a 1.500m solution of cacl2. How many grams of cacl2 are in the solution

Chemistry

1 answer:

Stella [2.4K]2 years ago

3 0

Answer:

120.7 g.

Explanation:

Molarity is defined as the no. of moles of a solute per 1.0 L of the solution.

M = (no. of moles of solute)/(V of the solution (L)).

∴ M = (mass/molar mass)of NaCl/(V of the solution (L)).

∴ mass of CaCl₂ = (M)(V of the solution (L))(molar mass) = (1.5 M)(0.725 L)(110.98 g/mol) = 120.7 g.

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Did Chelate compounds increase the rate of reaction?

Viefleur [7K]

Answer:

Chelate, any of a class of coordination or complex compounds consisting of a central metal atom attached to a large molecule, called a ligand, in a cyclic or ring structure. An example of a chelate ring occurs in the ethylenediamine-cadmium complex:

The ethylenediamine ligand has two points of attachment to the cadmium ion, thus forming a ring; it is known as a didentate ligand. (Three ethylenediamine ligands can attach to the Cd2+ ion, each one forming a ring as depicted above.) Ligands that can attach to the same metal ion at two or more points are known as polydentate ligands. All polydentate ligands are chelating agents.

Chelates are more stable than nonchelated compounds of comparable composition, and the more extensive the chelation—that is, the larger the number of ring closures to a metal atom—the more stable the compound. This phenomenon is called the chelate effect; it is generally attributed to an increase in the thermodynamic quantity called entropy that accompanies chelation. The stability of a chelate is also related to the number of atoms in the chelate ring. In general, chelates containing five- or six-membered rings are more stable than chelates with four-, seven-, or eight-membered rings.

Explanation:

7 0

2 years ago

What mass (g) of magnesium nitride (Mg3N2) can be made from the reaction of 1.22 g of magnesium with excess nitrogen? __Mg + __N

Citrus2011 [14]

<h3>Answer:</h3>

1.69 g Mg₃N₂

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Reading a Periodic Table

Stoichiometry

Using Dimensional Analysis
Reactions RxN

<h3>Explanation:</h3>

Step 1: Define

[RxN - Unbalanced] Mg + N₂ → Mg₃N₂

[RxN - Balanced] 3Mg + N₂ → Mg₃N₂

[Given] 1.22 g Mg

[Solve] grams Mg₃N₂

Step 2: Identify Conversions

[RxN] 3 mol Mg → Mg₃N₂

[PT] Molar Mass of Mg - 24.31 g/mol

[PT] Molar Mass of N - 14.01 g/mol

Molar Mass of Mg₃N₂ - 3(24.31) + 2(14.01) = 100.95 g/mol

Step 3: Stoich

[DA] Set up: $\displaystyle 1.22 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg})(\frac{1 \ mol \ Mg_3N_2}{3 \ mol \ Mg})(\frac{100.95 \ g \ Mg_3N_2}{1 \ mol\ Mg_3N_2})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 1.68873 \ g \ Mg_3N_2$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

1.68873 g Mg₃N₂ ≈ 1.69 g Mg₃N₂

8 0

2 years ago

Ammonium phosphate ((NH4)3 PO4) is an important ingredient to many fertilizers. It can be made by reacting phosphoric acid (H3 P

lara31 [8.8K]

Answer:

15.35 g of (NH₄)₃PO₄

Explanation:

First we need to look at the chemical reaction:

3 NH₃ + H₃PO₄ → (NH₄)₃PO₄

Now we calculate the number of moles of ammonia (NH₃):

number of moles = mass / molecular wight

number of moles = 5.24 / 17 = 0.308 moles of NH₃

Now from the chemical reaction we devise the following reasoning:

if 3 moles of NH₃ are produce 1 mole of (NH₄)₃PO₄

then 0.308 moles of NH₃ are produce X moles of (NH₄)₃PO₄

X = (0.308 × 1) / 3 = 0.103 moles of (NH₄)₃PO₄

mass = number of moles × molecular wight

mass = 0.103 × 149 = 15.35 g of (NH₄)₃PO₄

4 0

2 years ago

How does carbon dioxide behave differently (its properties) as a gas compared with as a solid?

ankoles [38]

It is effected by diffusion (the power of smell and wind spread) but a solid is not.

4 0

2 years ago

A 7.0 g sample of a hydrocarbon (a molecule that has only hydrogen and carbon) is subject to combustion analysis. The mass of CO

Akimi4 [234]

Answer: The empirical formula for the given compound is $CH_2$

Explanation:

The chemical equation for the combustion of compound having carbon and hydrogen follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of $CO_2=22.0g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 22.0 g of carbon dioxide, $\frac{12}{44}\times 22.0=6g$ of carbon will be contained.

For calculating the mass of hydrogen:

Mass of hydrogen = Mass of sample - Mass of carbon

Mass of hydrogen = 7.0 g - 6 g

Mass of hydrogen = 1.0 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{6g}{12g/mole}=0.5moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.0g}{1g/mole}=1.0moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.5 moles.

For Carbon = $\frac{0.5}{0.5}=1$

For Hydrogen = $\frac{1.0}{0.5}=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : C : H = 1 : 2

Hence, the empirical formula for the given compound is $C_{1}H_{2}=CH_2$

4 0

3 years ago