Getting Started Elite Medical Academy Online Health Courses... 2__ Survey: Week 12 Partic... B Homepage - St. Peters...prometic 4-- Screen Reader: SnC14(a q) + 2 N a2O(a q) → Sn O2(s) + 4 Na Cl(a q) Fe(CIO3)2(aq) + Na2CrO4(aq) → FeCrO4(s) + 2 NaCIO3(aq) Screen Reader: Pb(Cl 03)4(a q)+ 4 N a Cl(aq) → PbC14(s) + 4 N a Cl 03(aq) SnC14(aq) + 2 Na2O(aq) → SnO2 (s) + 4 NaCl(aq) 3 Screen Reader: F e(CI O3)2(a q) + N a2Cr 04(a q) → F e Cr 04 (s) + 2 Na Cl 03(a q) 3 Pb(ClO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaClO3(aq) _2_- Screen Reader: Pb(Cl 03)2(a q)+ 2 N a Cl(aq) → PbCl2(s) + 2N a CI O3(aq) 3 Pb(ClO3)4(aq) + 4 Na3N(aq) → Pb3N4(s) + 12 NaCIO3(aq) 11 - Screen Reader: 3 Pb(CI O3)4(a q)+ 4 N a3N(a q) → Pb3N4(s) + 12 N aCl O3(aq) 1. 3 2:4 3.6 4.8 Fundamentals of N 4 Match each reaction to the sum of the coefficients of its net ionic reaction. For example, sum of coefficients of 2 H2 + O2 → 2 H2O is 2 + 1 + 2 = 5. If you are having trouble with this exercis 2 Fe(CIO3)3(aq) + 3 Na2CrO4(aq) → Fe2(CrO4)3(s) + 6 NaCIO3(aq) __1__ 2 Screen Reader: 2 F e(CI O3)3(aq) + 3 N a2Cr 04(a q) → F e2(Cr 04)3(s) + 6 Na Cl O3(aq) Pb(ClO3)4(aq) + 4 NaCl(aq) → PbC14(s) + 4 NaCIO3(aq) 4 Screen Reader: Pb(Cl 03)4(a q)+ 4 N a Cl(aq) → PbC14(s) + 4 N a Cl 03(aq) SnC14(aq) + 2 Na2O(aq) → SnO2(s) + 4 NaCl(aq) Screen Reader: SnC14(a q) + 2 N a20(a q) → Sn O2(s) + 4 Na Cl(aq) Fe(CIO3)2(aq) + Na2CrO4(aq) → FeCrO4(s) + 2 NaCIO3(aq) 3 Screen Reader: Fe(CI O3)2(a q) + N a2Cr 04(a q) → F e Cr 04 (s) + 2 Na Cl O3(a q) 1.3 2.4 3.6 4.8 4
Getting Started Elite Medical Academy Online Health Courses... 2__ Survey: Week 12 Partic... B Homepage - St. Peters...prometic 4-- Screen Reader: SnC14(a q) + 2 N a2O(a q) → Sn O2(s) + 4 Na Cl(a q) Fe(CIO3)2(aq) + Na2CrO4(aq) → FeCrO4(s) + 2 NaCIO3(aq) Screen Reader: Pb(Cl 03)4(a q)+ 4 N a Cl(aq) → PbC14(s) + 4 N a Cl 03(aq) SnC14(aq) + 2 Na2O(aq) → SnO2 (s) + 4 NaCl(aq) 3 Screen Reader: F e(CI O3)2(a q) + N a2Cr 04(a q) → F e Cr 04 (s) + 2 Na Cl 03(a q) 3 Pb(ClO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaClO3(aq) _2_- Screen Reader: Pb(Cl 03)2(a q)+ 2 N a Cl(aq) → PbCl2(s) + 2N a CI O3(aq) 3 Pb(ClO3)4(aq) + 4 Na3N(aq) → Pb3N4(s) + 12 NaCIO3(aq) 11 - Screen Reader: 3 Pb(CI O3)4(a q)+ 4 N a3N(a q) → Pb3N4(s) + 12 N aCl O3(aq) 1. 3 2:4 3.6 4.8 Fundamentals of N 4 Match each reaction to the sum of the coefficients of its net ionic reaction. For example, sum of coefficients of 2 H2 + O2 → 2 H2O is 2 + 1 + 2 = 5. If you are having trouble with this exercis 2 Fe(CIO3)3(aq) + 3 Na2CrO4(aq) → Fe2(CrO4)3(s) + 6 NaCIO3(aq) __1__ 2 Screen Reader: 2 F e(CI O3)3(aq) + 3 N a2Cr 04(a q) → F e2(Cr 04)3(s) + 6 Na Cl O3(aq) Pb(ClO3)4(aq) + 4 NaCl(aq) → PbC14(s) + 4 NaCIO3(aq) 4 Screen Reader: Pb(Cl 03)4(a q)+ 4 N a Cl(aq) → PbC14(s) + 4 N a Cl 03(aq) SnC14(aq) + 2 Na2O(aq) → SnO2(s) + 4 NaCl(aq) Screen Reader: SnC14(a q) + 2 N a20(a q) → Sn O2(s) + 4 Na Cl(aq) Fe(CIO3)2(aq) + Na2CrO4(aq) → FeCrO4(s) + 2 NaCIO3(aq) 3 Screen Reader: Fe(CI O3)2(a q) + N a2Cr 04(a q) → F e Cr 04 (s) + 2 Na Cl O3(a q) 1.3 2.4 3.6 4.8 4
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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States of Matter
The substance that constitutes everything in the universe is known as matter. Matter comprises atoms which in turn are composed of electrons, protons, and neutrons. Different atoms combine together to give rise to molecules that act as a foundation for all kinds of substances. There are five states of matter based on their energies of attraction, namely solid, liquid, gases, plasma, and BEC (Bose-Einstein condensates).
Chemical Reactions and Equations
When a chemical species is transformed into another chemical species it is said to have undergone a chemical reaction. It consists of breaking existing bonds and forming new bonds by changing the position of electrons. These reactions are best explained using a chemical equation.
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Transcribed Image Text:Getting Started Elite Medical Academy Online Health Courses...
2__
Survey: Week 12 Partic... B Homepage - St. Peters...prometic
4-- Screen Reader: SnC14(a q) + 2 N a2O(a q) → Sn O2(s) + 4 Na
Cl(a q)
Fe(CIO3)2(aq) + Na2CrO4(aq) → FeCrO4(s) + 2 NaCIO3(aq)
Screen Reader: Pb(Cl 03)4(a q)+ 4 N a Cl(aq) → PbC14(s) + 4 N
a Cl 03(aq)
SnC14(aq) + 2 Na2O(aq) → SnO2 (s) + 4 NaCl(aq)
3 Screen Reader: F e(CI O3)2(a q) + N a2Cr 04(a q) → F e Cr 04
(s) + 2 Na Cl 03(a q)
3
Pb(ClO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaClO3(aq)
_2_- Screen Reader: Pb(Cl 03)2(a q)+ 2 N a Cl(aq) → PbCl2(s) + 2N
a CI O3(aq)
3 Pb(ClO3)4(aq) + 4 Na3N(aq) → Pb3N4(s) + 12 NaCIO3(aq)
11
-
Screen Reader: 3 Pb(CI O3)4(a q)+ 4 N a3N(a q) → Pb3N4(s) +
12 N aCl O3(aq)
1. 3
2:4
3.6
4.8
Fundamentals of N
4
Transcribed Image Text:Match each reaction to the sum of the coefficients of its net ionic reaction.
For example, sum of coefficients of 2 H2 + O2 → 2 H2O is 2 + 1 + 2 = 5. If you are having trouble with this exercis
2 Fe(CIO3)3(aq) + 3 Na2CrO4(aq) → Fe2(CrO4)3(s) + 6
NaCIO3(aq)
__1__
2
Screen Reader: 2 F e(CI O3)3(aq) + 3 N a2Cr 04(a q) → F
e2(Cr 04)3(s) + 6 Na Cl O3(aq)
Pb(ClO3)4(aq) + 4 NaCl(aq) → PbC14(s) + 4 NaCIO3(aq)
4
Screen Reader: Pb(Cl 03)4(a q)+ 4 N a Cl(aq) → PbC14(s) + 4 N
a Cl 03(aq)
SnC14(aq) + 2 Na2O(aq) → SnO2(s) + 4 NaCl(aq)
Screen Reader: SnC14(a q) + 2 N a20(a q) → Sn O2(s) + 4 Na
Cl(aq)
Fe(CIO3)2(aq) + Na2CrO4(aq) → FeCrO4(s) + 2 NaCIO3(aq)
3 Screen Reader: Fe(CI O3)2(a q) + N a2Cr 04(a q) → F e Cr 04
(s) + 2 Na Cl O3(a q)
1.3
2.4
3.6
4.8
4
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