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2 years ago

Fe(NO3)3 Percent composition

Chemistry

1 answer:

Scilla [17]2 years ago

5 0

Answer:

Fe: 23.1%

N: 17.4%

O: 59.5%

Explanation:

You take the grams that each weigh on the periodic table and add it all up. Afterward you create fractions of each over the whole. (Example: Fe/Fe+N+O) then you divide and turn that decimal into a percentage.

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A can was filled with crushed ice, sealed, and massed. The ice was melted by slowly warming the can and its contents. No water v

Lera25 [3.4K]

Answer:

A) The mass would be the same.

Explanation:

Since there is no loss of any particle to vapor during the phase change process from solid to liquid, the mass of the before and after the process will remain the same.

In this way, the law of conservation of mass is obeyed.
Mass is the amount of matter contained in a substance.
Since there is no room for escape or matter loss, the mass will remain the same.

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2 years ago

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What is the weight of dark matter

Ostrovityanka [42]

Answer:

Knowing this, researchers from the University of Southern Denmark decided to investigate the size of these hypothetical hidden particles. According to the team, dark matter could weigh more than 10 billion billion (10^9) times more than a proton.

Explanation:

If this is true, a single dark matter particle could weigh about 1 microgram, which is about one-third the mass of a human cell (a typical human cell weighs about 3.5 micrograms), and right under the threshold for a particle to become a black hole.

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3 years ago

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!: (a) 20 cm: 80

RoseWind [281]

Answer:

THE LENGTH OF THE AIR COLUMN IS 9.5 CM

Explanation:

Taking the atmospheric pressure to be 760 mmHg;

When the capillary tube is held horizontally, the pressure of the tube is 760 mmHg

when the capillary tube is held vertically, the pressure increases by 4 cm = 40 mm

The new pressure of the tube is hence, 760 + 40 mmHg = 800 mmHg

Using the pressure forlmula;

P1 V1 = P2 V2

P1 A1 L1 = P2 A2 L2

where A1 and A2 is the area of the capillary tube and it is equal, it cancels out.

P1 l1 = P2 l2

l2 = P1 l1 / P2

l2 = 760 * 10 / 800

l2 = 9.5 cm

The length of the air in the tube is 9.5 cm.

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3 years ago

Early historical models of the solar system were geocentric. Which of these phrases describes a geocentric solar system?

likoan [24]

Answer:

Explanation:

3 0

2 years ago

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Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water

madreJ [45]

Explanation:

The given data is as follows.

Temperature of dry bulb of air = $25^{o}C$

Dew point = $10^{o}C$ = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

Partial pressure of water vapor $(P_{a})$ = vapor pressure of water at a given temperature $(P^{s}_{a})$

Using Antoine's equation we get the following.

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}$

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}$

= 0.17079

$P^{s}_{a}$ = 1.18624 kPa

As total pressure $(P_{t})$ = atmospheric pressure = 760 mm Hg

= 101..325 kPa

The absolute humidity of inlet air = $\frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

$\frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

= 0.00735 kg $H_{2}O$ / kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg $H_{2}O$ / kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

0.07 kg - 0.00735 kg

= 0.06265 kg $H_{2}O$ for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

$0.06265 kg \times 100$

= 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0

3 years ago