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3 years ago

What properties does Chlorine have on its own?

Chemistry

1 answer:

natulia [17]3 years ago

4 0

Answer:

When isolated as a free element, chlorine takes the form of a greenish-yellow gas, which is 2.5 times heavier than air and smells like bleach. Chorine is the second-most-abundant halogen and the second-lightest halogen on Earth, after fluorine

Explanation:

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If the actual yield of sodium chloride from the reaction of

lions [1.4K]

If the actual yield of sodium chloride from the reaction of 8.3 g of sodium and 4.5 g of chlorine is 6.4 g, what is the percent yield?

Answer : 2Na(s) + Cl2(g) → 2NaCl(s)

Explanation: Friend me!!!

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2 years ago

Effectus [21]

Answer:

There are 5! goodluck,

Explanation:

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2 years ago

While you were "sweating" your chemistry test, water vapor evaporates from your body, absorbing 50,000 J of energy. (assume no

TiliK225 [7]

Answer:

$\boxed {\boxed {\sf B. \ 22 \ grams}}$

Explanation:

We need to use the formula for heat of vaporization.

$Q=H_{vap}*m$

Identify the variables.

The heat absorbed by the evaporating water is the <u>latent heat of vaporization. </u>For water, that is 2260 Joules per gram.

Q is the energy, in this problem, 50,000 Joules.

m is the mass, which is unknown.

$H_{vap}=2260 \ J/g\\Q=50,000 \ J \\$

Substitute the values into the formula.

$50,000 \ J=2260 \ J/g*m$

We want to find the mass. We must isolate the variable, m.

m is being multiplied by 2260 J/g. The inverse operation of multiplication is division. Divide both sides by 2260 J/g.

$\frac{50,000 \ J}{2260 \ J/g} =\frac{2260 \ J/g*m}{2260 \ J/g}$

$\frac{50,000 \ J}{2260 \ J/g} =m$

Divide. Note that the Joules (J) will cancel each other out.

$\frac{50,000 \ }{2260 \ g} =m$

$22.1238938 \ g =m$

Round to the nearest whole number. The 1 in the tenth place tells us to leave the number as is.

$22 \ g \approx m$

The mass is about 22 grams, so choice B is correct.

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3 years ago

Which phrase correctly describes the outer shell of electrons of each atom in most covalent compounds?

d1i1m1o1n [39]

A, I just took the quiz and that was the answer

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3 years ago

Read 2 more answers

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