Which green house gas is produced by commercial refrigeration and air conditioning systems

If an unknown element didplays many extremly strong metallic properties, where is it most likely to be located on the periodic t

Ipatiy [6.2K]

This is late but for anyone else who needs it...It's D. Far left

4 0

3 years ago

Anabolic reactions _______ bonds, whereas catabolic reactions __________ bonds. A. decrease; increase. B. break; make C. weaken;

nexus9112 [7]

Answer:

The corrext answer is E. make; break

Explanation:

In living organisms, the metabolism is either anabolic or catabolic where anabolic metabolism is energy consuming and catabolic metabolism is eneegy releasesing. It should however be noted that anabolic reaction builds or biosynthesize new mollecular structures while catabolic reaction breaks down complex structure bonds into simple structures

The braking down of bonds in catabolic reations realeses energy to sustain the anabolic rection process for the formation of new bonds

6 0

3 years ago

2. How many grams of water can be warmed from 25.0°C to 37.0°C by the addition of 8,064 calories?

sertanlavr [38]

Answer:

672 g

Explanation:

We can calculate the mass of water that can be warmed from 25.0°C to 37.0°C by the addition of 8,064 calories using the following expression.

$Q = c \times m \times \Delta T$

where,

c: specific heat of the water

m: mass

ΔT: change in the temperature

$m = \frac{Q}{c \times \Delta T } = \frac{8,064cal}{(1cal/g. \° C) \times (37.0 \° C - 25.0 \° C) } = 672 g$

The mass of water that can be warmed under these conditions is 672 grams.

5 0

3 years ago

During studies of the following reaction (i), a chemical engineer measured a less-than-expected yield of N2 and discovered that

bonufazy [111]

Answer:

Maximum expected yield = 87.2%

Explanation:

Equations of reactions:

Main reaction: N₂O₄(l) + 2N₂H₄(l) ---> 3N₂(g) + 4H₂O(g)

Side reaction: 2N₂O₄(l) + N₂H₄(l) ----> 6NO(g) + 2H₂O(g)

Molar mass of N₂O₄ = 92 g/mol; molar mass of N₂H₄ = 32 g/mol; molar mass of N₂ = 28 g/mol; molar mass of of NO = 30 g/mol; molar mass of water = 18 g/mol

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂.

101.1 g of N₂O₄ will react with 2 * 32 * 101.1 / 92 g of N₂H₄ = 70.33 g of N₂H₄

<em>N₂O₄ is the limiting reactant</em>

101.1 g of N₂O₄ will react to produce 3 * 14 * 101.1 / 92 g of N₂ = 46.15 g of N₂

In the side reaction, (6 * 30 g) of NO is produced from (2 * 92 g) of N₂O₄ and 32 g of N₂H₄

12.7 g of N₂O₄ will be produced from ( 2 * 92 * 12.7/180 g) of N₂O₄ and (32 * 12.7/180) g of N₂H₄ to produce

mass of N₂O₄ used = 12.98 g

mass of N₂H₄ used = 2.26 g

mass of N₂O₄ left for main reaction = 101.1 - 12.98 = 88.12 g

mass of N₂H₄ left for main reaction = 101.1 - 2.26 = 98.84 g

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂

88.12 g of N₂O₄ will react with 2 * 32 * 88.12 / 92 g of N₂H₄ = 61.30 g of N₂H₄

N₂O₄ is the limiting reactant.

88.12 g of N₂O₄ will to react produce 3 * 14 * 88.12 / 92 g of N₂ = 40.23 g of N₂

Percentage yield = (theoretical yield/actual yield) * 100%

Percentage yield = (40.23/46.15) * 100% = 87.2%

Therefore, maximum expected yield = 87.2%

4 0

3 years ago

Water (10 kg/s) at 1 bar pressure and 50 C is pumped isothermally to 10 bar. What is the pump work? (Use the steam tables.) O -7

11Alexandr11 [23.1K]

Explanation:

For an isothermal process equation will be as follows.

W = nRT ln $\frac{P_{1}}{P_{2}}$

It is given that mass is 10 kg/s or 10,000 g/s (as 1 kg = 1000 g). So, calculate number of moles of water as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{10000 g/s}{18 g/mol}$

= 555.55 mol/s

= 556 mol/s (approx)

As T = $50^{o}C$ or (50 + 273.15) K = 323.15 K. Hence, putting the given values into the above formula as follows.

W = nRT ln[/tex]\frac{P_{1}}{P_{2}}[/tex]

= $556 mol/s \times 8.314 J/ K mol K \times 323.15 K \times ln\frac{1}{10}$

= $556 mol/s \times 8.314 J/ K mol K \times 323.15 K \times -2.303$

= -3440193.809 J/s

Negative sign shows work is done by the pump. Since, 1 J = 0.001 kJ. Therefore, converting the calculated value into kJ as follows.

$3440193.809 J/s \times \frac{0.001 kJ}{1 J}$

= 3440.193 kJ/s

= 3451 kJ/s (approx)

Thus, we can conclude that the pump work is 3451 kJ/s.

6 0

3 years ago