1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
anyanavicka [17]
3 years ago
13

What adaptation might help a plant survive in an environment with cold winters?

Chemistry
2 answers:
Alenkinab [10]3 years ago
8 0
Having thin leaves to keep in more water (like like the pine tree :3)
emmasim [6.3K]3 years ago
7 0

Waxy, needlelike leaves. -Apex

You might be interested in
When substances are formed that are different from the starting substances, a physical change has occurred. True False
Sergio039 [100]
False, because physical change does not involve the formation of substances. An example of physical change is change of state e.g water freezing .It is still water if you were to melt it.
4 0
3 years ago
Read 2 more answers
Read the chemical equation.
Sladkaya [172]

Answer:- D. 1.8 moles of Fe and 2.7molCO_2 .

Solution:- The balanced equation is:

Fe_2O_3+3CO\rightarrow 2Fe+3CO_2

let's first figure out the limiting reactant using the given moles and mol ratio:

1.8molFe_2O_3(\frac{3molCO}{1molFe_2O_3})

= 5.4 mol CO

From calculations, 5.4 moles of CO are required to react completely with 1.8 moles of Iron(III)oxide but only 2.7 moles of CO are available. It means CO is limiting reactant.

Products moles depends on limiting reactant. Let's calculate the moles of each reactant formed for given 2.7 moles of CO.

2.7molCO(\frac{2molFe}{3molCO})

= 1.8 mol Fe

2.7molCO(\frac{3molCO_2}{3molCO})

= 2.7molCO_2

So, the correct choice is D.  1.8 moles of Fe and 2.7molCO_2 are formed.

6 0
3 years ago
How many grams of NaCl are needed to make 0.800 liter of a 5.00 M solution?
ziro4ka [17]

Answer:

             0.324 g is required to make 5.00 M solution of NaCl in 0.800 L.

Given data:

                 Molarity = 5.00 M

                 Formula Mass = 58.5 g/mol

                 Required volume = 0.800 L

To Find;

               Mass in gram = ?

Solution:

              Formula for calculating mass in gram is given as,

              Mass in gram = Molarity × Formula mass × Volume required / 1000 putting values

              Mass in gram = 5.00 M × 58.5 g/mol × 0.800 L / 1000

               Mass in gram = 0.234 g


6 0
3 years ago
Write a balanced equation for the neutralization reaction between the following (you do not need to include states of matter):
givi [52]

Answer:

Both have the same balance on a weight with. 2 dimensional figures on the area formula

Explanation:

8 0
2 years ago
PLz help ASAP
Andrews [41]
  1. it is at the bottom trust me i did the test
6 0
3 years ago
Read 2 more answers
Other questions:
  • A gas of 190 mL at a pressure of 74 atm can be expected to change its pressure when its volume changes to 30.0 mL. Express its n
    15·2 answers
  • Using the graph below, at what depth does the thermocline begin?
    7·2 answers
  • What is the major organic product obtained from the following sequence of reactions? PhCH2CHO PhCH2CH2CHO PhCH2CH2COOH PhCH2COOH
    10·1 answer
  • Potassium chlorate decomposes according to the following chemical equation:
    8·2 answers
  • If a snail with a red shell crossed with a white shell, what color would the children have shown in a punter square
    6·1 answer
  • Which atom is involved in giving your heat energy to beat
    13·1 answer
  • Name an alkene that would yield the alcohol above on hydration. (Submit a single name even if there is more than one correct ans
    12·1 answer
  • If a concentrated solution of acetic acid is 99.5 % HC 2 H 3 O 2 and has a density of 1.05 g/mL, what is the concentration of th
    9·1 answer
  • Somebody help me asap please giving brainliest
    7·1 answer
  • Identify the arrows that represent the process of cooling.<br> liquid<br> gas<br> solid
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!