A delicate plant embryo is protected by what?

Chemistry

2 answers:

sweet [91]3 years ago

6 0

The answer is the seed coat.

astraxan [27]3 years ago

3 0

The answer is seed coat the seed

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How to write half reaction for H2O2(aq) -> O2(g)

Ghella [55]

The overall reaction is as follows:
H2O2 + 2ClO2 => 2ClO2- + O2 + 2H+
Half cell reactions:
ClO2 + e- => ClO2-
H2O2 => O2 + 2H+ + 2e-

7 0

3 years ago

1 C3H8 + 5 O2 --> 3 CO2 + 4H20. If 1.5 moles of C3H8 react, how many

UNO [17]

Answer:

First confirm the reaction is balanced:

C3H8 + 5O2 --> 3CO2 + 4H20 (3 cabon - check; 8 hydrogen - check; 10 oxygen - check).

a) In the equation there is a 5:1 ratio between propane and oxygen. We also know that number of mole is proportional to pressure and volume. Since pressure is constant (STP) then the volume of O2 is 7.2 * 5 = 36 litres.

b) For a near ideal gas that PV = nRT (combined gas law). So for 7.2 litres propane we find n(propane) = 101.3 * 7.2/8.314*298 ~ 0.29 mole (using metric units throughout for simplicity).

There is a 1:3 ratio between propane and CO2. Therefore 3 * 0.29 = 0.87 mole of CO2 is produced.

MW(CO2) ~ 44 g/mol. Therefore m(CO2) = 44 * 0.87 ~ 38.3 g

c) We know we need more oxygen than propane (due to the 1:5 ratio) so oxygen is the limiting reagent. Again Volume is proportional to number of mole and we see there is a 5:4 ratio between oxygen and water. Therefore the volume of water vapour produced will be (4/5) * 15 = 12 litres.

The other questions use the same technique and will give you some much needed practice.

Explanation:

7 0

3 years ago

there are 125 lima beans 56 garbanzo beans 39 kidney beans and 144 green peas in a container. find the percent by number of gree

Sauron [17]

39.6% of the beans are green beans

5 0

3 years ago

The specific heat of a certain type of cooking oil is 1.75 J/(g⋅°C). How much heat energy is needed to raise the temperature of

Mademuasel [1]

Answer:

\large \boxed{\textbf{609 kJ}}

Explanation:

The formula for the heat absorbed is

q = mCΔT

Data:

m = 2.07 kg

T₁ = 23 °C

T₂ = 191 °C

C = 1.75 J·°C⁻¹g⁻¹

Calculations:

1. Convert kilograms to grams

2.07 kg = 2070 g

2. Calculate ΔT

ΔT = T₂ - T₁ = 191 - 23 = 168 °C

3. Calculate q

$q = \text{2070 g} \times 1.75 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 168 \,^{\circ}\text{C} = \text{609 000 J} = \textbf{609 kJ}\\\text{The heat energy required is }\large \boxed{\textbf{609 kJ}}$