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damaskus [11]
2 years ago
5

Describe one weather condition that would most likely occur along both fronts

Chemistry
1 answer:
bazaltina [42]2 years ago
4 0

Answer:usually cloudy with precipitation because the warm air along most fronts is usually sinking and cooling or causing water to evaporate or maybe sinking and warming causing water to evaporate ect

Explanation:

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What is the scientific name for the output variable
amm1812

Answer:

Coefficient

Explanation:

I am not that sure, but just wanted to help.

4 0
3 years ago
Which of the following statements is a true statement regarding a solution with [H1+] =1x10-5 M and [OH1-]= 1x10-9 M?
zheka24 [161]

Answer:

B. The [H1+] >[OH1-] and the solution is acidic

4 0
3 years ago
True or false(Internal forces act on the outside of an object, and external forces act within an object.)
kkurt [141]
False I’m not sure exactly
4 0
3 years ago
A newly discovered element, Z, has two naturally occurring isotopes. 90.3 percent of the sample is an isotope with a mass of 267
blsea [12.9K]
atomic mass=percentage of isotope a * mass of  isotope a + percentage of isotope b * mass of  isotope b+...+percentage of isotope n * mass of isotope n.

Data:
mass of isotope₁=267.8 u
percentage of isotope₁=90.3%

mass of isotope₂=270.9 u
percentage of isotope₂=9.7%

Therefore:

atomic mass=(0.903)(267.8 u)+(0.097)(270.9 u)=
=241.8234 u + 26.2773 u≈268.1 u

Answer: the mass atomic of this element would be 268.1 u
7 0
3 years ago
What is the limiting reagent when a 2.00 g sample of ammonia is mixed with 4.00 g of oxygen?​
UNO [17]

Answer:

Ammonia is limiting reactant

Amount of oxygen left  = 0.035 mol

Explanation:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Balance chemical equation:

4NH₃ + 3O₂     →     2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

                      NH₃          :            N₂

                        4             :             2

                      0.12           :           2/4×0.12 = 0.06

                      NH₃         :            H₂O

                        4            :             6

                        0.12       :           6/4×0.12 = 0.18

                       

                       O₂            :            N₂

                        3             :             2

                      0.125        :           2/3×0.125 = 0.08

                        O₂           :            H₂O

                        3              :             6

                        0.125       :           6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

                        NH₃          :             O₂

                           4            :              3

                           0.12       :          3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left  = 0.125 - 0.09 = 0.035 mol

3 0
3 years ago
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