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svetoff [14.1K]
3 years ago
5

What can you infer about the growth of the seed type pictured in the graphs below? ( Note, the vertical axis represents the numb

er of seeds )
The graph shows Time in days from 0 to 5 on the x-axis and 0 to 100 on the y-axis with 5 percent light. Day 0 there are 100 seeds and 0 sprouts. The number of seeds and sprouts remains the same through 2.5 days. Then the number of seeds decreases while the number of sprouts increases. Day 5 there are 30 seeds and 70 sprouts.The graph shows Time in days from 0 to 5 on the x-axis and 0 to 100 on the y-axis with 80 percent light. Day 0 there are 100 seeds and 0 sprouts. The number of seeds and sprouts remains the same through 2.5 days. Then the number of seeds decreases while the number of sprouts increases. Day 5 there are 30 seeds and 70 sprouts.
a.
More seeds grow with more light.
b.
More seeds grow with less light.
c.
The seeds grow faster with more light.
d.
Light has no effect on seed growth.
Chemistry
1 answer:
liraira [26]3 years ago
5 0

Answer:

D

Explanation:

I just did it on Edge

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While 1 gram of fat provides 9 calories, 1 gram of glucose provides 4 calories. Why is that?
seropon [69]

Answer:

fat always has more calories than glucose does

Explanation:

hope hope this helps!

6 0
3 years ago
Strike anywhere matches contain the compound tetraphosphorus trisulfide, which burns to form tetraphosphorus decaoxide and sulfu
erica [24]

Answer:

194.6 mL of SO₂

Explanation:

The reaction that takes place is:

P₄S₃ + 6O₂(g) → P₄O₁₀ + 3SO₂(g)

<u>To solve this problem we need to use PV=nRT</u>, so first let's convert the given units:

  • 23.8 °C → 23.8 + 273.15 = 296.95 K
  • 747 torr → 747/760 = 0.983 atm

We need to calculate V, so in order to do that we calculate n, using the mass of the reactant (P₄S₃):

0.576 g P₄S₃ * \frac{1molP_{4}S_{3}}{220gP_{4}S_{3}} *\frac{3molSO_{2}}{1molP_{4}S_{3}} = 7.85 * 10⁻³ mol SO₂ = n

  • Now we calculate V:

PV=nRT

0.983 atm * V =  7.85 * 10⁻³ mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 296.95 K

V = 0.1946 L

  • Finally we convert L into mL:

0.1946 * 1000 = 194.6 mL

8 0
3 years ago
the element carbon has two common isotopes: C-12 (12 U) AND C-13 (13.003355 U). IF THE AVERAGE ATOMIC MASS OF CARBON IS 12.0107
oksian1 [2.3K]

Answer:

The percent isotopic abundance of C- 12 is 98.93 %

The percent isotopic abundance of C- 13 is 1.07 %

Explanation:

we know there are two naturally occurring isotopes of carbon, C-12 (12u)  and C-13 (13.003355)

First of all we will set the fraction for both isotopes

X for the isotopes having mass 13.003355

1-x for isotopes having mass 12

The average atomic mass of carbon is 12.0107

we will use the following equation,

13.003355x + 12 (1-x) = 12.0107

13.003355x + 12 - 12x = 12.0107

13.003355x- 12x = 12.0107 -12

1.003355x = 0.0107

x= 0.0107/1.003355

x= 0.0107

0.0107 × 100 = 1.07 %

1.07 % is abundance of C-13 because we solve the fraction x.

now we will calculate the abundance of C-12.

(1-x)

1-0.0107 =0.9893

0.9893 × 100= 98.93 %

98.93 % for C-12.

7 0
3 years ago
formic acid buffer containing 0.50 M HCOOH and 0.50 M HCOONa has a pH of 3.77. What will the pH be after 0.010 mol of NaOH has b
HACTEHA [7]

Answer:

pH = 3.95

Explanation:

It is possible to calculate the pH of a buffer using H-H equation.

pH = pka + log₁₀ [HCOONa] / [HCOOH]

If concentration of [HCOONa] = [HCOOH] = 0.50M and pH = 3.77:

3.77 = pka + log₁₀ [0.50] / [0.50]

<em>3.77 = pka</em>

<em />

Knowing pKa, the NaOH reacts with HCOOH, thus:

HCOOH + NaOH → HCOONa + H₂O

That means the NaOH you add reacts with HCOOH producing more HCOONa.

Initial moles of 100.0mL = 0.1000L:

[HCOOH] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOOH

[HCOONa] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOONa

After the reaction, moles of each species is:

0.0500moles HCOOH - 0.010 moles NaOH (Moles added of NaOH) = 0.0400 moles HCOOH

0.0500moles HCOONa + 0.010 moles NaOH (Moles added of NaOH) = 0.0600 moles HCOONa

With these moles of the buffer, you can calculate pH:

pH = 3.77 + log₁₀ [0.0600] / [0.0400]

<h3>pH = 3.95</h3>

3 0
3 years ago
A sample of 0.800 mol of nitrogen has a pressure of 0.600 atm and a
schepotkina [342]

Answer:

274K

Explanation:

5 0
3 years ago
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