When the equation H202 + H2O + O2 is completely balanced, the SUM of all the coefficients will be

MULTIPLE CHOICE<br> What is the molality (m) of 3.5 mol of sugar in 2.5 kg of water?

otez555 [7]

The molality of the solution is 1.4 moles per kilogram (molal).

The molality of the solution is equal to the number of moles of sugar (solute) divided by the number of kilograms of water (solvent). Hence, the molality is determined below:

$m = \frac{3.5\,mol}{2.5\,kg}$

$m = 1.4\,\frac{mol}{kg}$

The molality of the solution is 1.4 moles per kilogram (molal).

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5 0

3 years ago

Assume the volume of the crew cabin is 74,000 L. The pressure is maintained at around 1.00 atm, ideally with an 80% nitrogen and

hjlf

Answer:

The number of moles of oxygen present in the crew cabin at any given time is 615.309 moles

Explanation:

The given parameters are;

Volume of the crew cabin = 74,000 L

Pressure of the crew cabin = 1.00 atm

Percentage of nitrogen in the mixture of gases in the cabin = 80%

Percentage of oxygen in the mixture of gases in the cabin = 20%

Temperature of the cabin = 20°C = 293.15 K

Therefore, volume of oxygen in the crew cabin = 20% of 74,000 L

Hence, volume of oxygen in the crew cabin = $\frac{20}{100} \times 74,000 \, L = 14,800 \, L$

From the universal gas equation, we have;

$n = \frac{P \times V}{R \times T}$

Where:

n = Number of moles of oxygen

P = Pressure = 1.00 atm

V = Volume of oxygen = 14,800 L

T = Temperature = 293.15 K

R = Universal Gas Constant = 0.08205 L·atm/(mol·K)

Plugging in the values, we have;

$n = \frac{1 \times 14,800 }{0.08205 \times 293.15 } = 615.309 \, moles$

The number of moles of oxygen present in the crew cabin at any given time = 615.309 moles.

3 0

4 years ago

Copper reacts with nitric acid releasing nitrogen dioxide gas. Why is this answer<br>right?

Contact [7]

Explanation:

Nitric acid reacts with copper according to the reaction: ... 4 HNO3(l) + Cu(s) ==> Cu(NO3)2(s and aq) + 2 NO2(g) + 2 H2O(l) The copper nitrate salt that forms is a deep blue color. The nitrogen dioxide is a maroon vapor.

4 0

4 years ago

A typical fast food meal consists of a burger, fries, and a soft-drink and contains 58.0 grams of fat, 39.0 grams of protein, an

Burka [1]

Answer:

371.12 minutes would it take to completely burn off the meal.

Explanation:

Amount of fat in the food = 58.0 g

The respective fuel values for fat = 38 kJ/g

Amount of protein = 39.0 g

The respective fuel values for protein = 17 kJ/g

Amount of carbohydrates in the food = 177 g

The respective fuel values for carbohydrates = 17 kJ/g

Total energy derived from the meal = Q

$Q=58.0 g\times 38 kJ/g+39.0 g\times 17 kJ/g+177 g\times 17 kJ/g$

Q = 5876 kJ

Energy used while jogging for 1 hour = 950.0 kJ

Energy used while jogging for 60 minutes = 950.0 kJ

Let the time require to burn 5876 kJ of energy be t.

So, $\frac{950.0 kJ}{60 min}=\frac{5876 kJ}{t}$

$\frac{5876 kJ\times 60 min}{950.0 kJ}=371.12 min$

371.12 minutes would it take to completely burn off the meal.

5 0

3 years ago

How many milliliters of nitrogen, N2, would have to be collected at 99.19 kPa and 28oC to have a sample containing 0.015 moles o

Semmy [17]

Answer:

378mL

Explanation:

The following data were obtained from the question:

Pressure (P) = 99.19 kPa

Temperature (T) = 28°C

Number of mole (n) = 0.015 mole

Volume (V) =...?

Next, we shall convert the pressure and temperature to appropriate units. This is illustrated below:

For Pressure:

101.325 KPa = 1 atm

Therefore, 99.19 kPa = 99.19/101.325 = 0.98 atm

For Temperature:

T(K) = T(°C) + 273

T(°C) = 28°C

T(K) = 28°C + 273 = 301K.

Next we shall determine the volume of N2. The volume of N2 can be obtained by using the ideal gas equation as shown below:

PV = nRT

Pressure (P) = 0.98 atm

Temperature (T) = 301K

Number of mole (n) = 0.015 mole

Gas constant (R) = 0.0821atm.L/Kmol.

Volume (V) =...?

0.98 x V = 0.015 x 0.0821 x 301

Divide both side by 0.98

V = (0.015 x 0.0821 x 301) /0.98

V = 0.378 L

Finally, we shall convert 0.378 L to millilitres (mL). This is illustrated below:

1L = 1000mL

Therefore, 0.378L = 0.378 x 1000 = 378mL

Therefore, the volume of N2 collected is 378mL

4 0

3 years ago