PLEASEEEEEE answer these 2 questions!! I WILL GIVE BRAINLIEST AND POINTS

An unknown element is determined to have 22 neutrons and a mass number of 40.

Amiraneli [1.4K]

It’s Calcium hope that helps!

7 0

3 years ago

If NaCl has a mass of 3.2g, what is the volume of chlorine gas at STP?

Shalnov [3]

Hey there!

Molar mass NaCl = 58.44 g/mol

58.44 g ----------------- 22.4 ( at STP )

3.2 g -------------------- Volume ??

Volume = ( 3.2 x 22.4 ) / 58.44

Volume = 71.68 / 58.44

Volume = 1.226 L

Hope this helps!

5 0

3 years ago

Read 2 more answers

an isotope of cesium (cesium-137) has a half-life of 30 years if 1.0g of cesium-137 disintegrates over a period of 90 years how

Juliette [100K]

The answer is after 3 half lives

4 0

3 years ago

The current global population is 7.7 billion people, and the WHO estimates the minimum water need per person per day is 15 L. An

anygoal [31]

Answer:

a) The amount of freshwater available as groundwater, lakes and rivers, does not even reach one day the need for consumption for the current global population. ( t = 5.489 E-13 day )

b) The annual terrestrial precipitation, reaches to sustain the drinking water needs for the current global population

Explanation:

Let P = 7.7 billion people = 7.7 E12 person

∴ water needed for the total population for one day:

⇒ water amount = 7.7 E12 person * ( 15 L / person. day ) = 1.155 E14 L H2O /day

⇒ water amount = 1.155 E14 L H2O/day * 1 E1 Km² H2O / L H2O = 1.155 E15 Km² H2O/day * ( 365 day / year ) = 4.216 E17 Km²/year

∴ freshwater available:

freshwater = 6.34 E2 Km² H2O

how long will this water sustain the current population?

⇒ t = 6.34 E2 Km² * day / 1.155 E15 Km² = 5.489 E-13 day

this amount of freshwater does not even meet the need of the current global population.

∴ the annual terrestrial precipitation (Py) = 505000 Km³/year..........from literature

⇒ Py = 505000 Km³ H2O/year * ( 1000 m/Km )³ * ( 1000L/m³ )

⇒ Py = 5.05 E17 L H2O/year * ( 1 E1 Km² / L ) = 5.05 E18 Km² H2O/year

⇒ Py > water amount

the annual terrestrial precipitation of water, reaches to sustain the drinking water needs.

4 0

3 years ago

An unknown compound contains only carbon, hydrogen, and oxygen (CxHyOz). Combustion of 6.50 g of this compound produced 9.53 g o

OverLord2011 [107]

Answer:

There were 0.216 moles of oxygen in the sample

Explanation:

Step 1: Data given

Mass of compound = 6.50 grams

Mass of CO2 = 9.53 grams

Molar mass of CO2 = 44.01 g/mol

Mass of H2O = 3.90 grams

Molar mass H2O = 18.02 g/mol

Step 2: Calculate moles CO2

Moles CO2 = 9.53 grams / 44.01 g/mol

Moles CO2 = 0.217 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.217 moles CO2 we have 0.217 moles C

Step 4: Calculate mass C

Mass C = 0.217 moles C *12.01 g/mol

Mass C = 2.61 grams

Step 5: calculate moles H2O

Moles H2O = 3.90 grams / 18.02 g/mol

Moles H2O = 0.216 moles

Step 6: Calculate moles H

In 1 mol H2O we have 2 moles H

In 0.216 moles H2O we have 0.433 moles H

Step 7: Calculate mass H

Mass H = 0.433 moles * 1.01 g/mol

Mass H = 0.437 grams

Step 8: Calculate mass O

Mass O = 6.50 grams - 2.61 grams - 0.437 grams

Mass O = 3.453 grams

Step 9: Calculate moles O

Moles O = 3.453 grams / 16.0 g/mol

Moles O = 0.216 moles

There were 0.216 moles of oxygen in the sample

4 0

3 years ago