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3 years ago

What is a major difference between Animalia and Plantae kingdoms?

Chemistry

1 answer:

iren [92.7K]3 years ago

5 0

Kingdom Plantae

1. Photosynthesis occurrence due to the presence of chloroplast.

2. Are immobile.

3. Can grow throughout their life.

Kingdom Animalia

1.Since they are non-green in nature and do not produce chlorophyll, they do not undergo photosynthesis.

2.Are mobile.

3.Achieve a maximum size and then stop growing.

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KatRina [158]

Answer A,They can form a triple covalent bond.

Explanation:

7 0

3 years ago

What is the chemical name of the compound Cu(NO3)2? Use the list of polyatomic ions and the periodic table to help you answer.

vlabodo [156]

Answer:

C. Copper (II) nitrate

Explanation:

It's an inorganic compound that forms a blue crystalline solid

8 0

3 years ago

Hydrazine reacts with oxygen according to the following equation: N2H4(g) +O2(g) → N2(g) + 2 H2O(l) How many L of N2, measured a

mihalych1998 [28]

Answer:

V ≈ 646.50 L

General Formulas and Concepts:

Chemistry - Gas Laws

Reading a Periodic Table
Stoichiometry
Combined Gas Law: PV = nRT
R constant - 62.4 (L · torr)/(mol · K)
Kelvin Conversion: K = °C + 273.15

Explanation:

Step 1: Define

RxN: N₂H₄ (g) + O₂ (g) → N₂ (g) + 2H₂O (l)

Given: 34.9 °C, 755.08 torr, 914.894 g H₂O

Step 2: Identify Conversions

Kelvin Conversion

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Step 3: Convert

Stoichiometry: $914.894 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O} )(\frac{1 \ mol \ N_2}{2 \ mol \ H_2O} )$ = 25.3955 mol N₂

Temperature: 34.9 + 273.15 = 308.05 K

Step 4: Find Volume

Substitute variables: (755.08 torr)V = (25.3955 mol)(62.4 (L · torr)/(mol · K))(308.05 K)
Multiply: (755.08 torr)V = 488160 L · torr
Isolate V: V = 646.502 L

Step 5: Check

We are given 5 sig figs as our lowest. Follow sig fig rules and round.

646.502 L ≈ 646.50 L

6 0

3 years ago

A 1.2516 gram sample of a mixture of caco3 and na2so4 was analyzed by dissolving the sample and completely precipitating the ca

Dennis_Churaev [7]

Answer:

0.009725 moles of H2C2O4

0.009725 moles CaCO3

Mass percentage = 77.77%

Explanation:

Step 1: The balanced equation

2MnO4- +5C2H2O4+6H+ →2Mn2+ +10CO2+8H2O

We can see that for 2 moles of Mno4- consumed , there is 5 moles of C2H2O4 needed and 6 moles H+ to produce 2 moles Mn2+, 10 moles of CO2 and 8 moles of H2O

Step 2: Calculate moles of MnO4-

Molarity = Moles/volume

Moles of Mno4- = Molarity of MnO4- * Volume of Mno4-

Moles of Mno4- = 0.1092M * 35.62 *10^-3 L

Moles of MnO4- = 0.00389 moles

Step 3: Calculate moles of H2C2O4

Since there is needed 5 moles of C2H2O4 to consume 2 moles of MnO4-

then for 0.00389 moles of MnO4-, there is 5/2 *0.00389 = 0.009725 moles of H2C2O4

Step 4: Calculate moles of CaCO3

moles of H2C2O4 = moles CaCO3, therefore, 0.009725 moles H2C2O4 = 0.009725 moles CaCO3

Step 5: Calculate mass of CaCO3

Molar mass of CaCO3 = 100.09 g/mole

Mass of CaCO3 = moles of CaCO3 * Molar mass of CaCO3

Mass of CaCO3 = 0.009725 moles * 100.09 g/mole = 0.9734 g

Step 6: Calculate percentage by weight of CaCO3

Mass of CaCO3 = 0.9734g

Mass of original sample = 1.2516g

Mass percentage = 0.9734/1.2516 *100% = 77.77%

6 0

3 years ago

Journal making expressing qualitative data and quantitative data.

barxatty [35]

Journal making expressing data and quantitative data

6 0

3 years ago