Which of these facts best illustrates why regulation of alcohol consumption is necessary

A chalk company gets an order for 3000 boxes of chalk at the beginning of the school year, the largest order they have ever rece

VMariaS [2998]

The chalk production's efficiency is noted to be 82 %

To determine this efficiency, we utilize the formula,

% Efficiency $=\frac{Actual yield (g)}{Theoretical yield (g)} *100$

Finding the actual yield based on % efficiency and the theoretical yield:

82 % $=\frac{Actual yield}{400,000 g} * 100$

Actual yield = $\frac{82 *400000 g}{100} = 32,800 g$

Each box contains 145 grams of chalk

Total boxes = $32800 g *\frac{1 box}{145 g} = 2262 boxes$

The chalk box company aims to produce 3000 boxes. However, with 82 % efficiency, they can only produce 2262 boxes, thus falling short of their target.

8 0

3 months ago

A substance is 35.7% carbon by mass. How much carbon could be recovered from 769 g of the substance? 1. 180 mol 2. 258 mol 3. 22

KiRa [2933]

Response:

22.9 moles (Option 5)

Clarification:

A substance being 35.7% carbon by mass signifies that in every 100 g of the substance, there is 35.7 g of carbon.

Using a rule of three, we get:

100 g of substance ___ has __ 35.7 g of C

769 g of substance ___ has ___ ( 769.35.7) / 100 = 274.5 g of C

1 mol of C = 12 g/m

Mass / Molar mass = Moles

274.5 g / 12 g/m = 22.9 moles

4 0

2 months ago

A 19.3-g mixture of oxygen and argon is found to occupy a volume of 16.2 l when measured at 675.9 mmhg and 43.4oc. what is the p

KiRa [2933]

<span>The partial pressure of oxygen is 438.0 mmHg. The ideal gas equation is expressed as PV = nRT where P represents pressure, V denotes volume, n is the number of moles, R is the ideal gas constant (8.3144598 (L*kPa)/(K*mol)), and T signifies absolute temperature. To convert from Celsius to Kelvin, we have 43.4 + 273.15 = 316.55 K. For the pressure conversion from mmHg to kPa: 675.9 mmHg * 0.133322387415 = 90.11260165 kPa. When solving for n using the ideal gas equation, we derive n = PV / (RT) which provides n = 90.11260165 kPa * 16.2 L / (8.3144598 (L*kPa)/(K*mol) * 316.55 K)= 1459.824147 L*kPa / 2631.94225 (L*kPa)/(mol), resulting in n = 0.554656603 mol. Thus, we have 0.554656603 moles of gas particles. Next, we determine the contribution from oxygen. The atomic weight of oxygen is 15.999 g/mol, while argon is 39.948 g/mol, and the molar mass of O2 is 31.998 g/mol. We establish the relationships where M is the number of moles of O2, and 0.554656603 - M gives the number of moles of Ar. Setting up the equation: M * 31.998 + (0.554656603 - M) * 39.948 = 19.3, we solve for M resulting in 0.359424148 moles of oxygen out of 0.554656603 total moles. This leads to oxygen providing 0.359424148 / 0.554656603 = 0.648012024 or 64.8012024% of the total pressure of 675.9 mmHg. The partial pressure therefore calculates to 675.9 * 0.648012024 = 437.9913271 mmHg, rounded to 438.0 mmHg</span>

7 0

3 months ago