5. Which pair of elements would combine to forin an ionic compound?

PLZ HELP ME What are the five shared characteristics of all living things?

djverab [1.8K]

Answer:

D

Explanation:

I believe the answer is D.

5 0

3 years ago

Sylvanite is a mineral that contains 28.0% gold by mass. How much sylvanite would you need to dig up to obtain 73.0 g of gold?

Scorpion4ik [409]

You would have to dig up 261 g of sylvanite.

Mass of sylvanite = 73.0 g Au × (100 g sylvanite/28.0 g Au) = <em>261 g</em> sylvanite.

3 0

3 years ago

Antoine Lavoisier burned metals in sealed jars. What is your prediction of his experimental results?

Aliun [14]

Answer:

The mass of the jar and contents remained the same after the metal was burned.

Explanation:

My prediction about the experimental results is that the mass of the jar and contents remained the same after the metal was burned in the jar.

This is compliance with the law of conservation of mass which states that in a chemical reaction, matter is neither created nor destroyed by bonds are rearranged for new compounds to form.

In compliance with this law, it is expected that the mass of the jar and its content will remain the same before and after the reaction.
No new material was added and no material was removed from the jar.

8 0

3 years ago

Is there law of conversation of Volume? If two different liquids say you take 100ml of liquid A and 100 ml of liquid B and mix t

victus00 [196]

Answer:

200 mL

Explanation:

You simply add the volumes :) If this was a case that involved titration, you would use the formula M1V1 = M2V2. I do not know if that is what you were referring to but based on the information you gave, you simply add the two volumes.

4 0

3 years ago

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