Answer:
Oversight of weights and measures ensures correct evaluations of goods and services so that everyone receives a fair exchange in the marketplace. It also acts as a deterrent, promoting honesty among traders.
Explanation:
Answer:
The temperature of the gas rises.
Explanation:
This is classified as an ISOCHORIC process where the volume remains unchanged. There is no work done by the system.
The gas only receives internal energy from the heat transferred to it from the surroundings.
In this situation, the pressure also increases.
Response:
The mass percentage of a solution comprising 7.6 grams of sucrose and 83.4 grams of water equals 8.351 %.
Details:
Provided data:
Sucrose mass = 7.6 grams
Water mass = 83.4 grams
In this scenario, sucrose acts as the solute, while water is the solvent.
The calculation for mass percent of a solution is done using the following formula:
Mass percent = (Mass of Solute/Mass of Solution)(100)
As sucrose is the solute, the mass equals 7.6 grams.
The total mass of the solution, which includes both sucrose and water, comes out to:
Total mass = 7.6 grams + 83.4 grams = 91 grams
Therefore, applying the values gives mass percent = (7.6/91)(100) = 8.351 %.
The result is 200 g. Given that the molar mass of CaCl2 is 110.98 g/mol, this indicates that there are 110.98 g in 1 L of a 1 M solution. Let's calculate the amount of CaCl2 in 0.720 M. Using the proportion 110.98 g: 1 M = x: 0.720 M, we find x to be 79.90 g. Therefore, in 1 L of a 0.720 M solution, there is 79.90 g. Next, we need to create ten beakers with 250 mL each, totaling 10 * 250 mL = 2500 mL or 2.5 L. Then, using the equation 79.90 g: 1 L = x: 2.5 L, we calculate x = 79.90 g * 2.5 L: 1 L, resulting in x = 199.75 g, approximately 200 g.
Context:
175 kilograms of methane (CH4) is to be converted into hydrogen cyanide (HCN)
The equation that balances this reaction is listed here:
2 CH4<span> + 2 NH</span>3<span> + 3 O</span>2<span> → 2 HCN + 6 H</span>2<span>O
</span>
To find the quantities of ammonia and oxygen needed, we will use 175 kg of CH4 as our reference.
Molar masses are as follows:
CH4 = 16 kg/kmol
NH3 = 17 kg/kmol
O2 = 32 kg/kmol
For ammonia: mass of NH3 = 175 kg CH4 / 16 kg/kmol * (2/2) * 17 kg/kmol
This results in 185.94 kg of NH3 required
For oxygen: mass of O2 = 175 kg CH4 / 16 kg/kmol * (3/2) * 32 kg/kmol
So the mass of O2 needed equals 525 kg
To derive the mass of oxygen: mass of O = 525 kg / 32 kg/kmol * (1/2) * 16 kg/kmol
This gives a mass of O equal to 131.25 kg O
