balance the following equation:kclo3 (s) —heat→ kc...
balance the following equation:kclo3 (s) —heat→ kcl (s) + o2 (g)
kclo3(s) — heat→ kcl (s) + o2(g)      reactant        productk        1                   1cl       1                   1o       3                    2[readmore]2 kclo3(s) — heat→ 2 kcl (s) + 3 o2(g)      reactant        productk        2                   2cl       2                   2o       6                    6the balanced equation for the reaction is 2 kclo3(s) — heat→ 2 kcl (s) + 3 o2(g)
solution: `ch\_(4)(g) + 2o\_(2)(g) to co\_(2)(g) + 2h\_(2)o(g) + "heat. (evolved)"` AgNO3(aq) + NaCl(s) = AgCl(s) + NaNO3  
n(AgCl) = 0.765 g / M(AgCl)  
M(AgCl) = 108 + 35.5 = 143.5 g/mol  
n(AgCl) = 0.765 g /143.5 g/mol = 0.00533 mol  
n(AgCl) = chloride ions = 0.00533 mol  
C(chloride ions) = n/V  
C(chloride ions) = 0.00533 mol/1l = 0.00533 mol/ L (a)  
m(NaCl)=n(NaCl)*M(NaCl)  
n(NaCl) = 0.00533 mol  
M(NaCl) = 23 + 35.5 = 58.5 g/mol  
m(NaCl) = 0.00533 mol * 58.5 g/mol = 0.3115 g (b)  
  
**Answer:**  
**a) 0.00533 mol/ L**  
**b) 0.3115 g** a) 3ag + 4hn03 > 3agno3 + no + 2h2o

b) mno2 + 4hcl > 2h2o + mncl2 + cl2 i. given:valency of carbon is 4valency of chlorine is 1by applying criss-cross method:formula: ccl4thus, the formula for carbon tetrachloride is ccl4.ii. given:valency of element x is 1valency of oxygen is 2by applying criss-cross method:formula: x2o y = 3x

It can be observed that x = −1, y = −3 and x = 1, y = 3 are solutions of the above equation. Therefore, the solution table is as follows.



| x | -1 | 1 |
| y | -3 | 3 |

The graph of the above equation is constructed as follows.

![](https://www.sarthaks.com/?qa=blob&qa_blobid=15681428413536225923) f = ma(t) = kte^(-at) => a(t) = k/m * te^(-at)

v(t) = ∫a(t) dt = k/(ma^2) * (-ate^(-at) - e^(-at) + c)
= -k/(ma^2) * e^(-at)*(at + 1 + ce^(at)) action of cl2 on potassium bromide solution: chlorine reacts with kbr, to form potassium chloride, displacing bromine. on sapling its

oxidation:
cu --> cu^2+ + 2e^-

reduction:
o\_2 + 4e^- -->2o^2- given, x3 – 1, x – 1 as pair of polynomialsneed to find the quotient and remainder⇒ to find the quotient and remainder the given equation can be written as p(x) = (x – a)q(x) + b⇒ since, the polynomial is of third degree we can write the q(x) as x2 + ax + b∴ p(x) = (x – a)( x2 + ax + b) + c⇒ x3 + 1 = (x + 1)(x2 + ax + b) + c⇒ x3 + 1 = (x3 + x2 + ax2 + ax + bx + b) + c⇒ x3 + 1 = x3 + (a + 1)x2 + (b + a)x + (c + b)∴ a + 1 = 0, b + a = 0, c + b = 1⇒ a = – 1, b = 1⇒ c + b = 1⇒ c = 0quotient = x2 + ax + b = x2 – x + 1remainder = 0 NaBr   

  

SO3   

  

CaCl2   

  

Mg3N2