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What is the density of a testbook in science?
A text book has a density near 0.09 lb/in3 or 2530 kg/m3.
How much mass would a large book have?
The mass of a large book can vary depending on its size and number of pages. On average, a large hardcover book might weigh around 1-2 pounds (0.45-0.9 kg).
What are some things that weigh 1 kg?
It has been said that an average middle school text book weighs about 1 kilogram.
What is the morality solution that has 3 mol of glucose in 6 kg if water?
To calculate the molality (m) of a solution, you use the formula: ( m = \frac{\text{moles of solute}}{\text{kg of solvent}} ). In this case, with 3 moles of glucose in 6 kg of water, the molality would be ( m = \frac{3 , \text{mol}}{6 , \text{kg}} = 0.5 , \text{mol/kg} ). Therefore, the molality of the solution is 0.5 mol/kg.
What is the molality of a solution that has 6 mol of CACI2 on 3 kg of water?
Molality (m) is calculated using the formula ( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} ). For a solution with 6 moles of CaCl₂ dissolved in 3 kg of water, the molality would be ( m = \frac{6 \text{ mol}}{3 \text{ kg}} = 2 \text{ mol/kg} ). Therefore, the molality of the solution is 2 mol/kg.
What is the molality of a solution that has 6 mol of cac2 in 3 kg of water?
To calculate the molality (m) of a solution, use the formula: [ m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} ] In this case, there are 6 moles of calcium carbide (CaC₂) and 3 kg of water. Thus, the molality is: [ m = \frac{6 , \text{mol}}{3 , \text{kg}} = 2 , \text{mol/kg} ] Therefore, the molality of the solution is 2 mol/kg.
What is the molality of a solution that has 6 mol of CaCI2 in 3kg of water?
Molality (m) is calculated using the formula ( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} ). In this case, with 6 moles of CaCl₂ dissolved in 3 kg of water, the molality would be ( m = \frac{6 , \text{mol}}{3 , \text{kg}} = 2 , \text{mol/kg} ). Therefore, the molality of the solution is 2 mol/kg.
What is the molality of a solution that has 3 mol of glucose in 6kg of water?
Molality (m) is calculated using the formula ( m = \frac{\text{moles of solute}}{\text{kg of solvent}} ). In this case, there are 3 moles of glucose and 6 kg of water. Therefore, the molality of the solution is ( m = \frac{3 , \text{mol}}{6 , \text{kg}} = 0.5 , \text{mol/kg} ). Thus, the molality of the solution is 0.5 mol/kg.
What is the molality of a solution made by a dissolving 2 moles of naoh in 6 kg of water Apex?
Molality (m) is calculated using the formula ( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} ). In this case, with 2 moles of NaOH dissolved in 6 kg of water, the molality is ( m = \frac{2 , \text{moles}}{6 , \text{kg}} = \frac{1}{3} , \text{mol/kg} ). Therefore, the molality of the solution is approximately 0.33 mol/kg.
Spaceship 1 and Spaceship 2 have equal masses of 300 kg. Spaceship 1 has a speed of 0 ms and Spaceship 2 has a speed of 4 ms. They collide and stick together. What is their speed?
To find the final speed after the collision, we can use the principle of conservation of momentum. The initial momentum of the system is the momentum of Spaceship 2, since Spaceship 1 is at rest: ( p_{initial} = m_2 \times v_2 = 300 , \text{kg} \times 4 , \text{m/s} = 1200 , \text{kg m/s} ). After the collision, the combined mass is ( 300 , \text{kg} + 300 , \text{kg} = 600 , \text{kg} ). Setting the initial momentum equal to the final momentum, we have ( 1200 , \text{kg m/s} = 600 , \text{kg} \times v_{final} ), which gives ( v_{final} = 2 , \text{m/s} ).
What would you need to do to calculate the molality of 0.2 kg of NaCI in 3 kg of?
To calculate the molality of 0.2 kg of NaCl in 3 kg of water, first determine the number of moles of NaCl by using its molar mass (approximately 58.44 g/mol). Convert 0.2 kg of NaCl to grams (200 g) and calculate the moles: ( \text{moles} = \frac{200 \text{ g}}{58.44 \text{ g/mol}} ). Finally, use the formula for molality, which is moles of solute per kilogram of solvent: ( \text{molality} = \frac{\text{moles of NaCl}}{3 \text{ kg of water}} ).
A 2600-kg truck runs into the rear of a 1000-kg car that was stationary. The truck and car are locked together after the collision and move with speed 8 ms. What was the speed of the truck before the?
To find the speed of the truck before the collision, we can use the law of conservation of momentum. The initial momentum of the system (truck + car) before the collision must equal the final momentum after the collision. The equation is: [(m_{truck} \cdot v_{truck}) + (m_{car} \cdot v_{car}) = (m_{truck} + m_{car}) \cdot v_{final}] Given that the car is stationary, (v_{car} = 0), and substituting the known values: [(2600 , \text{kg} \cdot v_{truck}) + (1000 , \text{kg} \cdot 0) = (2600 , \text{kg} + 1000 , \text{kg}) \cdot 8 , \text{m/s}] This simplifies to: [2600 , \text{kg} \cdot v_{truck} = 3600 , \text{kg} \cdot 8 , \text{m/s}] Calculating gives: [2600 , \text{kg} \cdot v_{truck} = 28800 , \text{kg m/s}] Thus, [v_{truck} = \frac{28800 , \text{kg m/s}}{2600 , \text{kg}} \approx 11.08 , \text{m/s}] Therefore, the speed of the truck before the collision was approximately 11.08 m/s.
