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Paraphin [41]
3 years ago
15

Which is common to the life cycles of both flowering and non-flowering plants?

Chemistry
2 answers:
andrew11 [14]3 years ago
7 0
Answer: STAY DORMANT
Explanantion: because staying dormant is a common circle of flowering and non flowering
Stolb23 [73]3 years ago
4 0

Answer:

2 stay dormant

bbdndndn

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How many molecules are in 2.00 moles of h2o
Sedaia [141]
1 mole = 6.022×10^23 atoms. 1 water molecule = 2 Hydrogen atoms + 1 oxygen atom. So, 1 mole H2O = 1.2044×10^24 hydrogen atoms. Therefore 2 mole H2O will have 2.4088×10^24 hydrogen atoms.
8 0
3 years ago
A certain substance X has a normal freezing point of -6.4 C and a molal freezing point depression constant Kf= 3.96 degrees C.kg
Brut [27]

Answer:  1.0\times 10^2g

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=(-6.4-(13.6))^0C=7.2^0C = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte like urea)

K_f = freezing point constant = 3.96^0C/m

m= molality

\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}}\times \text{weight of solvent in kg}}

Weight of solvent (X)= 950 g = 0.95 kg  

Molar mass of non electrolyte (urea) = 60.06 g/mol

Mass of non electrolyte (urea) added = ?

7.2=1\times 3.96\times \frac{xg}{60.06 g/mol\times 0.95kg}

x=1.0\times 10^2g

Thus 1.0\times 10^2g urea was dissolved.

8 0
4 years ago
4. Find the pH at each of the following points in the titration of 25 mL of 0.3 M HF with 0.3 M NaOH. The Ka value is 6.6x10-4 a
yawa3891 [41]

Explanation:

Since HF is a weak acid, the use of an ICE table is required to find the pH. The question gives us the concentration of the HF.

HF+H2O⇌H3O++F−HF+H2O⇌H3O++F−

Initial0.3 M-0 M0 MChange- X-+ X+XEquilibrium0.3 - X-X MX M

Writing the information from the ICE Table in Equation form yields

6.6×10−4=x20.3−x6.6×10−4=x20.3−x

Manipulating the equation to get everything on one side yields

0=x2+6.6×10−4x−1.98×10−40=x2+6.6×10−4x−1.98×10−4

Now this information is plugged into the quadratic formula to give

x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√2x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)2

The quadratic formula yields that x=0.013745 and x=-0.014405

However we can rule out x=-0.014405 because there cannot be negative concentrations. Therefore to get the pH we plug the concentration of H3O+ into the equation pH=-log(0.013745) and get pH=1.86

6 0
3 years ago
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