Phototropism is a type of tropism in which a plant or plant part responds to light. According to this question, a student wanted to investigate the effect of light on the growth of cress seedlings. The student used three different pots for the experiment.

Pot 1 was placed with light from above. Pot 2 was placed in a cupboard with no light. Pot 3 was placed in a window with light from one direction only. However, the image attached to this question shows that the plants in the different pots face different directions in response to light, which depicts phototropism

5 0

3 years ago

Chem help what's the answer to 8 and why will give brainliest

laiz [17]

avogadros law states that under constant temp n pressure, volume of a gas is directly proportional to amount of gas

amount of gas1 = 352millimoles

amount of gas2 = 352+100 = 452millimoles

new vol = old vol x 452/352 = 25.2 x 452/352

= 32.4 mL

4 0

3 years ago

Read 2 more answers

1 Na2CO3(aq) + 1 CaCl2(aq) → 1 CaCO3(s) + 2 NaCl(aq) 4. Use the balanced chemical equation from the last question to solve this

LenKa [72]

<h3>Answer:</h3>

0.6 g NaCl

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)

[Given] 0.5 g Na₂CO₃ reacted with excess CaCl₂

<u>Step 2: Identify Conversions</u>

[RxN] Na₂CO₃ → 2NaCl

Molar Mass of Na - 22.99 g/mol

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol

Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 0.5 \ g \ Na_2CO_3(\frac{1 \ mol \ Na_2CO_3}{105.99 \ g \ Na_2CO_3})(\frac{2 \ mol \ NaCl}{1 \ mol \ Na_2CO_3})(\frac{58.44 \ g \ NaCl}{1 \ mol \ NaCl})$
Multiply/Divide: $\displaystyle 0.551373 \ g \ NaCl$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

0.551373 g NaCl ≈ 0.6 g NaCl

5 0

2 years ago

What type of tide is produced when earth is directly between the sun and the moon?

Yuki888 [10]

High tide or Neap tides which are normally really high.

4 0

3 years ago