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Darina [25.2K]
3 years ago
6

A student has a 2.123 g sample of a hydrated salt. she heats it to drive off all the water. after heating to constant mass, the

salt weighs 1.861 g. the anhydrous salt is copper(ii) phosphate. what is the empirical formula of the hydrated salt?
Chemistry
1 answer:
morpeh [17]3 years ago
4 0
The mass of hydrated salt - 2.123 g
mass of anhydrous salt - 1.861 g
mass that has been reduced is the mass of water that has been heated and lost from the compound thereby making the salt anhydrous.
therefore mass of water lost - 2.123 - 1.861 = 0.262 g
number of moles of water lost - 0.262 g / 18 g/mol = 0.0146 mol 
number of moles of salt - 1.861 g / 380.6 g/mol = 0.00490 mol 
molar ratio of moles of water to moles of salt 
molar ratio = 0.146 mol / 0.00490 mol = 2.98 rounded off to 3
for every 1 mol of salt there are 3 moles of water
therefore empirical formula - Cu₃(PO₄)₂.3H₂O


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Calculate the value of Kp at 227°C for the equilibrium: 3 A(g) ⇌ B(g) + D(g) Kc = 5.15
horrorfan [7]
D is the answer I guess
3 0
2 years ago
Which is least likely to be reduced? <br> A. Zn^2+ <br> B. Fe^3+ <br> C. Cu^2+ <br> D. Fe^2+
alekssr [168]

A. Zn²⁺

<h3>Further explanation</h3>

Given

Cations of several elements

Required

The least to be reduced

Solution

If we look at the voltaic series:

<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe²⁺-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Fe³⁺-Ag-Pt-Au </em>

The electrode which is easier to reduce than the hydrogen (H2) electrode has a positive sign (E red= +) and is located to the right of the voltaic series (right of H)

The electrode which is easier to oxidize than the hydrogen (H2) electrode and is difficult to experience reduction has a negative sign (E red= -) and is located to the left of the voltaic series (left of H)

Or you can look at the standard reduction potential value of the metals in the answer options, and the most negative reduction E° value which will be difficult to reduce.

The Zn metal is located far left of the other metals in the answer choices, so it is the most difficult to reduce

6 0
2 years ago
I NEED HELP PLEASE, THANKS! :)
krok68 [10]

Answer:

\large \boxed{\text{2.20 g Pb}}

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:       239.27   32.00        207.2

            2PbS   +   3O₂   ⟶  2Pb   +   2SO₃

m/g:      2.54        1.88

2. Calculate the moles of each reactant

\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}

3. Calculate the moles of Pb from each reactant

\textbf{From PbS:}\\\text{Moles of Pb} =  \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol  Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}

4. Calculate the mass of Pb

\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}

5 0
2 years ago
Consider the equation: 2 nan3 (s) 2 na (s) + 3 n2. what is the mass of nan3 needed to produce 17.2 l nitrogen at stp?
Lerok [7]
The  mass  of NaN3  needed  to produce 17.2  L  nitrogen   at  STP is  calculated as  follows

find the  moles of N2   produced  at  STP
At  STP  1mole  of  gas =  22.4  L ,  what about  17.2 L  of  nitrogen

by cross  multiplication

= (1 mole x17.2  L)/ 22.4 L= 0.768  moles

2NaN3 =2Na  +3 N2
by   use of  mole ratio  between NaN3  to N2  (2:3) the  moles  of NaN3  =  0.768  x2/3 =  0.512  moles  of NaN3

mass of NaN3  is therefore =moles of NaN3  xmolar  mass of NaN3

=0.512moles  x 65 g/mol =33.28 grams  of NaN3



3 0
3 years ago
Which reaction describes an alpha emission?
Tcecarenko [31]

88226Ra→ 86222Rn + 24He is the reaction that describes an alpha emission because radiations are released.

<h3>Describes an alpha emission?</h3>

Alpha radiation occurs when the nucleus of an atom becomes unstable and alpha particles are released in order to restore stability. Alpha decay occurs in elements that have high atomic numbers, such as uranium, radium, and thorium etc.

So we can conclude that 88226Ra→ 86222Rn + 24He is the reaction that describes an alpha emission because radiations are released.

Learn more about radiation here: brainly.com/question/893656

#SJ1

7 0
2 years ago
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