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Alona [7]
3 years ago
9

Which household items would be considered a base? Select all that apply. *

Chemistry
1 answer:
mrs_skeptik [129]3 years ago
6 0

Answer:

Drain Cleaner, soap, bleach

i think thats it, sorry if i got something wrong

You might be interested in
The orbital period of planet Venus is 0.62 years. What is its distance from the sun?
Vikki [24]

Kepler's third law shows the relationship between the orbital period of an object and the distance between the object and the object it orbits.

The simplified version of this law is: P^2 = a^3

Where,

P = period of the orbit in years = 0.62 years

a = average distance from the object to the object it orbits in AU. The astronomical unit AU is a unit of length which is roughly equivalent to the distance from Earth to the Sun.

Therefore calculating for a:

0.62 ^ 2 = a ^ 3

a = 0.62 ^ (2/3)

a = 0.727 AU = 0.72 AU

Therefore we can interpret this as: The distance from Venus to the Sun is about 72% of the distance from Earth to Sun.

<span>Answer: B. 0.72 AU</span>

5 0
3 years ago
Photosynthesis by land plants leads to the fixation each year of about 1 kg of carbon on the average for each square meter of an
Rama09 [41]

Answer:

a) mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b) all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

Explanation:

first we calculate the moles of carbon

moles = mass/molar mass

= 1kg/12gmol⁻¹

= 1000g/12gmol⁻¹

= 83.33 mol

now using the ideal gas equation

we find the volume of co₂required based on 83.33 moles

PVco₂ = nRT

Vco₂ = nRT/P

Vco₂ = (83.22mol × 0.0821L atm k⁻¹ mol⁻¹ 298 K) / 1 atm

Vco₂ = 2083.73 L

so since CO₂ in air is 0.0390% by volume in the atmosphere, we find the the total amount of air required to obtain 1kg carbon

therefore

Vair × 0.0390/100 = 2038.73L

Vair = (2038.73L × 100) / 0.0390

Vair = 5.23 × 10⁶L

therefore 5.23 × 10⁶ L of air will be required to obtain 1kg carbon

a)

Here we calculate the mass of air over 1 square meter of surface.

Remember that atmospheric pressure is the consequence of the force exerted by all the air above the surface; 1 bar is equivalent to 1.020×10⁴kgm⁻²

NOW

mass of air = 1.020×10⁴kgm⁻² × 1m²

= 1.020×10⁴kg

= 1.020×10⁷g    [1kg = 10³g]

we now find the moles of air associated with it

moles = mass/molar mass

= 1.020 × 10⁷g / ( 20%×Mo₂ + 80%×Mn₂)

= 1.020 × 10⁷g / ( 20%×32gmol⁻¹ + 80%×28gmol⁻¹)

= 1.020 × 10⁷g / 28.8 gmol⁻¹

= 354166.67mol

so based on the question, for each mole (air), there is 0.0390% of CO₂

now to calculate the moles of CO₂ we say;

MolesCo₂ = 0.0390/100 × 354166.67mol

= 138.125 moles

Now we calculate mass of CO₂ from the above findings

Moles = mass/molar mass

mass = moles × molar mass

= 138.125 moles × 12gmol⁻¹

= 1657.5g

we covert to KG

= 1657.5g / 1000

mass = 1.65kg

therfore mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b)

to find the number years required to use up all the CO₂, WE SAY

Number of years = total carbon per m² of the forest / carbon used up per m² from the forest per year

Number of years = 1.65kgm⁻² / 1kg²year⁻¹

Number of years = 1.65 years

Therefore all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

6 0
3 years ago
Need help with this chem question thanks
IRISSAK [1]

Answer:

The answer to your question is letter A.

Explanation:

Isomers are molecules that have the same molecular formula but have a different structure. The molecule from which are looking an isomer has 5 carbons and 1 double bond. Then we need to look for another molecule with these components.

A.- This molecule has 5 carbons and 1 double bond, This structure is an isomer of the first one.

B.- This molecule has 3 carbons and 1 double bond, it's not an isomer of the first structure.

C. This molecule has 4 carbons and 1 triple bonds, it's not an isomer of the first structure.

D. This molecule has 5 carbons but it doesn't have any double bond, then it's not an isomer of the first structure.

5 0
3 years ago
A golf pro has 2100 J of kinetic energy and swings his driver which weighs .75 kg. What is the speed of his swing?
jeka57 [31]

Answer : The correct answer is 74.83 m/s .

The kinetic energy is energy possessed by any mass which is moving or have some speed . It is product of mass and velocity . It is expressed as :

KE = \frac{1}{2}  * m* v^2

Where KE = kinetic energy in J or kg\frac{m^2}{s^2} m = mass in Kg v = speed in m/s²

Unit of KE is Joules (J) .

Givne : KE = 2100 J mass = 0.75 kg v = ?

Plugging value in KE formula =>

2100 J = \frac{1}{2}  * 0.75 Kg * v^2

2100 J = 0.375 Kg * v^2

Dividing both side by 0.375 kg =>

\frac{2100 J}{0.375 Kg}  = \frac{0.375 Kg}{0.375 Kg } * v^2

v^2 = 5600 \frac{m^2}{s^2}

v= 74.83 \frac{m}{s}

8 0
3 years ago
Read 2 more answers
Te second period at the bottom of the table is called
katovenus [111]

Answer:

The second row elements are called Actinidies . abbreviated as Ac

Explanation:

5 0
3 years ago
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