yes
but in ac ckts complex form is used
To find the molality of the KCl solution, we can use the freezing point depression formula: (\Delta T_f = K_f \cdot m \cdot i), where (i) is the van 't Hoff factor (which is 2 for KCl). The freezing point depression (\Delta T_f) is 3°C (from 0°C to -3°C). Using (K_f = 1.86 , \text{°C kg/mol}), we can rearrange the formula to find molality (m): [ m = \frac{\Delta T_f}{K_f \cdot i} = \frac{3}{1.86 \cdot 2} \approx 0.805 , \text{mol/kg} ] Thus, the molality of the KCl solution is approximately 0.805 mol/kg.
To dissolve 91.6 g of KCl in 200 g of water, the water should ideally be at a temperature around 60°C. KCl has a moderate solubility in water, and heating the water increases the solubility of many salts, including KCl. However, the exact temperature can vary depending on the desired solubility and the specific conditions. It’s recommended to check solubility tables for precise values at different temperatures.
To calculate the freezing point of a 0.743 M aqueous solution of KCl, we first determine the van 't Hoff factor (i) for KCl, which dissociates into two ions: K⁺ and Cl⁻. Therefore, ( i = 2 ). Using the freezing point depression formula, (\Delta T_f = i \cdot K_f \cdot m), where ( K_f ) for water is approximately 1.86 °C kg/mol, we find (\Delta T_f = 2 \cdot 1.86 \cdot 0.743 \approx 2.77 °C). Thus, the freezing point of the solution is approximately (0 - 2.77 = -2.77 °C).
If lamp C burns out in a circuit where lamps A and B are connected in series with lamp C, then both lamps A and B will also go out. This is because the circuit is incomplete when lamp C fails, preventing current from flowing through lamps A and B. In a parallel circuit, however, if lamp C burns out, lamps A and B would remain lit as they have alternative paths for current. The specific behavior depends on how the lamps are connected in the circuit.
Yes. Example: .................................................... ...A * ........................................... ......|.\ ......................................... eg Euler circuit: ACDCBA ......|...\ ........... --------- ............. ......|.....\........./...............\............ The Hamilton circuit is impossible as it has two ......|.......\...../...................\.......... halves (ACD & CD) connected to each other only ......|.........\./.......................\........ at vertex C. Once vertex C has been reached in ......|.......C *........................* D.... one half, it can only be used to start a path in ......|........./.\......................./......... the other half, or complete the cycle in the ......|......./.....\.................../........... current half; or if the path starts at C, it will end ......|...../.........\.............../............. without the other half being visited before C is ......|.../ ........... --------- .............. revisited. ......|./ ........................................... ...B *.............................................. ......................................................
At 40°C, KBr is less soluble than KCl. At 80°C, the trend may reverse, with KCl possibly being less soluble than KBr. The solubility of salts typically increases with temperature, so KBr may be the least soluble at both 40°C and 80°C.
Approximately 34 grams of KCl can dissolve in 100g of water at 30°C.
The solubility of KCl in water at 80°C is approximately 380 g/L. To form a saturated solution, you would need to dissolve 380 g of KCl in 1 L (1000 g) of water. Since you are using 200 g of water, you would need 76 g of KCl to form a saturated solution.
This is a solution of 10 g KCl/100 g water.
The solubility of KCl at 100 degrees Celsius is approximately 56.5 g/100 mL of water. This means that at this temperature, 100 mL of water can dissolve 56.5 grams of KCl before reaching saturation.
Approximately 42 grams of KCl can dissolve in 100 grams of water at 80°C.
1. Identify (a) through (d) as reactants or products. Type your answers in the spaces provided: (a) (b) (c) (d) KOH + HCl KCl + H2O
The specific conductance of 0.1 N KCl solution is approximately 1.25 mS/cm at 25°C.
The vapor pressure of KCl depends on temperature. At room temperature (around 25°C), the vapor pressure of KCl is very low, close to negligible. As the temperature increases, the vapor pressure of KCl also increases, following the general trend for solids.
K2CO3 is the molecular formula for potassium carbonate. K2CO3 has a melting point of 1164 degrees Kelvin and is 138.21 grams per mole.
Infinite dilution molar conductivity is the limiting value of the molar conductivity as the concentration tends to zero. The molar conductivity of KCl at infinite dilution is approximately 149.9 S cm² mol⁻¹ at 25°C.
Vat is not applicable. because it is a interstate sale. so only cst is applicable . cst 2%. along with form 'c'...if form 'c' is not provided by the buyer, then state vat is applicable...