=> (7.9860 x 14.00) / (154.03 - 65.1)
According to BODMAS rule. ( Bracket, Of, Division, Multiplication, Addition and Subtraction) and based on significant rules.
=> (111.8) / 88.9
=> 1.2575
=> 1.26 ( rounded to Three significant figures)
The solubility of Cr(OH)3 in water is 1.26 x 10-8 M at 25 °C. Calculate the Ksp value for Cr(OH)3. A. 2.48 x 10-32 B. 1.98 x 10-24 C. 1.58 x 10-16 D. 1.26 x 10-8 E. 6.80 x 10-31
final volume of 2.7 L. B) 0.75 atm C) 1.3 atm D) 0.27 atm 4.3 x 1S4 15) What is the volume occupied by 35.2 g of methane gas (CH4) at 25°C and 1.0 atm? (R = 0.08206 L . atm/K·mol) A) 11.2 L B) 4.5 L C) 53.6L D) 49.2 L
Question 7 The solubility of Cr(OH)3 in water is 1.26 x 10-8 M at 25 °C. Calculate the Ksp value for Cr(OH)3. A. 6.80 x 10-31 B. 1.98 x 10-24 C.1.26 x 10-8 OD. 1.58 x 10-16 E. 2.48 x 10-32 > A Moving to another question will save this response.
1.3 Given x[n]-a-2?Mn + 2]-2Mn] + (1+ J)Mn-1] , find a) Ren] b)Imxnl c) Even^x[n] d) Oddxinl e)Im Evenfn] f)Even lms»in)
Part A Determine the derivative d of y222 + 1.3. C 15 xA 2+44*x 152 +44 You are correct. Your receipt no. is 165-4585 Part B Evaluate your expression by calculating the value of for3. 3.00 You are correct. Your receipt no. is 165-3986 Part C with respect to x of y- 5z3 + 22x2 + 1.3. The value of the integral at z-0 is-1.5D Submit Answer Incorrect. Tries 3/6 Previous Tries Part D Evaluate your expression by calculating the...
Evaluate the function for the given values. k(x) = (2x +2] (a) (b) k(-2.3) k(1.3) (d) نہ سما
1.3 Convert: (a) 16.3 m to mm (b) 16.3 m to km (c) 4 x 10-6 uF (microfarad) to pF (picofarad) (d) 2.3 ns to us (e) 3.6 x 10 v to MV (f) 0.03 mA (milliamp) to LA
46. What is the H30' concentration in 0.0013 M LiOH(aq) at 25 °C? (Kw = 1.01 x 10-4) a. 7.7 x 10-12 M b. 1.3 x 10-3 M c. 1.0 × 10-14 M. d. 1.0 x 10-7M e. 1.3 * 10-17 M
The molar solubility of Ag2S is 1.26 x 10-16 M in pure water. Calculate the Ksp for Ag25. A) 6.81 x 10-63 B) 1.12 * 10-8 C) 3.78 x 10-12 D) 8.00 * 10-48 E) 1.59 x 10-32 nun
If 500 mL of 1.3 x 10-6 M AgNO3 is mixed with 500 mL of 1.3 x 10-6 M NaBr, what will occur? For AgBr, Kp = 5 x 10-13 a) The concentration of Agt will be 1.3 x 10-6 M. b) Sodium bromide will precipitate. O c) 6.5 x 10-7 mol of AgBr will form. d) Silver(1) bromide will precipitate. e) No precipitation will oc occur