How many moles of Oxygen are in 0.52 mol of Ca(NO3)2?

slader An experiment is carried out where 13.9 g of solid NaOH is dissolved in 250.0 g of water in a coffee-cup calorimeter. Dis

kvv77 [185]

Answer:

The mass of the surrounding is $M_t = 263.9 \ g$

Explanation:

From the question we are told that

The mass of $NaOH$ is $m = 13.9 \ g$

The mass of water is $m_w = 250.0g$

The chemical equation for the dissociation process is

$NaOH _{(s) } ---> Na^{+}_{(aq)} + OH^{-} _{(aq)}$

The specific heat capacity of the mixture is $c_p = 4.18 J g^{-1} C^{-1}$

The combined mass of the solution is

$M_t = 13 + 250$

$M_t = 263.9 \ g$

The mass of the surround here is the mass of the coffee-cup calorimeter and this contain the mixture ( water and the NaOH ) so the mass of the surrounding is

$M_t = 263.9 \ g$

3 0

3 years ago

How do the atomic number and the mass number of an element

Doss [256]

Answer:

A.

Explanation:

3 0

2 years ago

Given the following unbalanced equation:

Lorico [155]

<h3>Answer:</h3>

8.01 mol MgO

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Moles
Compounds

<u>Stoichiometry</u>

Using Dimensional Analysis
Analyzing Reactions RxN

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] Mg + O₂ → MgO

[RxN - Balanced] 2Mg + O₂ → 2MgO

[Given] 8.01 moles Mg

[Solve] moles MgO

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol Mg → 2 mol MgO

<u>Step 3: Stoich</u>

[DA] Set up: $\displaystyle 8.01 \ mol \ Mg(\frac{2 \ mol \ MgO}{2 \ mol \ Mg})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 8.01 \ mol \ MgO$

3 0

2 years ago

An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.

Tju [1.3M]

Answer:

a) $\Delta S$

b) entropy of the sistem equal to a), entropy of the universe grater than a).

Explanation:

a) The change of entropy for a reversible process:

$\delta S=\frac{\delta Q}{T}$

$\Delta S=\frac{Q}{T}$

The energy balance:

$\delta U=[tex]\delta Q- \delta W$

If the process is isothermical the U doesn't change:

$0=[tex]\delta Q- \delta W$

$\delta Q= \delta W$

$Q= W$

The work:

$W=\int_{V1}^{V2}P*dV$

If it is an ideal gas:

$P=\frac{n*R*T}{V}$

$W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV$

Solving:

$W=n*R*T*ln(V2/V1)$

Replacing:

$\Delta S=\frac{n*R*T*ln(V2/V1)}{T}$

$\Delta S=n*R*ln(V2/V1)}$

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

$\Delta S$

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.

7 0

3 years ago

Oxygen gas, generated by the reaction 2KClO3(s)---2KCl(s)+3O2(g), is collected over water at 27•C in 3.72L vassel at a total pre

Julli [10]

Answer:

moles = 0.093 moles

Explanation:

In this case, we know that this reaction is taking plave in a vessel that has a 730 torr of total pressure.

The total pressure is a value obtained by:

Pt = Pwater + PO2

We need to know the pressure of O2, because then, with stoichiometry, we can calculate the moles of KClO3

The pressure of oxygen is:

PO2 = 730 - 26 = 704 Torr

Now, this pressure is in Torr, and we need to convert it to Atm, so:

704 Torr / 760 Torr = 0.9263 atm

Now, let's use the ideal gas equation:

PV = nRT

With this expression, we will calculate the moles of O2, and then, the moles of KClO3:

n = PV/RT

R = 0.082 L atm /K mol

P = 0.9263 atm

V = 3.72 L

T = 27 + 273 = 300 K

Replacing the data:

n = 0.9263 * 3.72 / 300 * 0.082

n = 0.14 moles

Finally, by stoichiometry, we know that 2 moles of KClO3 produces 3 moles of O2, so:

moles of KClO3 = 0.14 * 2/3 = 0.093 moles of KClO3

6 0

2 years ago