All categories

4 years ago

Is a seed a living thing?why

Chemistry

2 answers:

Alenkinab [10]4 years ago

7 0

Yes, because it wil grow and become a bigger seed or object

STALIN [3.7K]4 years ago

6 0

It has the potential for life when given the means, and it can die when it isn't given enough care, so I would say that it is.

You might be interested in

How much heat is released if 7.15 g cao(s) is added to 152 g of h2o(l)? cao(s) + h2o(l) → ca(oh)2(s) δh°rxn = –64.8 kj/mol?

aniked [119]

First, we calculate the number of moles of each reactant using the formula:

Moles = mass / molecular weight

CaO:
Moles = 7.15/56 = 0.128

Water:
Moles = 152/18 = 8.44

The reaction equation shows that the reactants must be present in an equal number of moles, so CaO will be the limiting reactant and 0.128 mole of calcium hydroxide will form.

The energy released is given by:

Heat of reaction * number of moles
= -64.8 * 0.128
= -8.29 kJ

8.29 kJ of energy will be released

6 0

4 years ago

What is e cell (in units of volts) for a galvanic cell constructed from barium and manganese electrodes?

gulaghasi [49]

Main Answer: The standard electrode potential of galvanic cell constructed from barium and manganese electrode is 4.097.

Explanation:

When the galvanic cell constructed between Manganese and Barium, the half reactions will be as follows:

Mn2+(aq) + 2e-⟶ Mn(s) E1 = −1.185

Ba2+(aq) + 2e-⟶ Ba(s) E2 = −2.912

By considering the above two reactions, the standard electrode potential can be calculated.

The galvanic cell standard electrode potential is given by E1 + E2 = 1.185 + 2.912 = 4.097 volts

What is standard electrode potential?

Standard electrode potential is defined as the measure of reducing power of any compound or element. Its units are volts.

To know more about galvanic cell, please visit:

brainly.com/question/13031093

#SPJ4

3 0

2 years ago

Jellyfish are only found in the deep ocean?<br><br>True or False

Ierofanga [76]

False. Jellyfish are found all over in the ocean

5 0

2 years ago

3.87 moles of iron bromide yielded10.4 moles of sodium bromide

Roman55 [17]

Answer: 89.57 %

<em>Hope it helps</em>

7 0

4 years ago

Determine the freezing point of a water solution of fructose (C6H12O6) made by dissolving 92.0 g of fructose in 202 g of water.

Naya [18.7K]

Answer:

THE FREEZING POINT OF A WATER SOLUTION OF FRUCTOSE MADE BY DISSOLVING 92 g OF FRUCTOSE IN 202g OF WATER IS -4.70 ◦C

Explanation:

To calculate the freezing point of a water solution of fructose,

1. calculate the molar mass of Fructose

( 12 * 6 + 1*12 + 16*6) =72 + 12 +96 = 180 g/mol

2. calculate the number of moles of fructose in the solution

number of moles = mass / molar mass

n = 92 g / 180 g/mol

n = 0.511 moles.

3. calculate the molarity of the solution

molarity = moles / mass of water in kg

molarity = 0.5111 / 202 g /1000 g

molarity = 0.5111 / 0.202

molarity = 2.529 M

4. calculate the change in the freezing point of pure solvent and solution ΔTf

ΔTf = Kf * molarity of the solute

Kf = 1.86 ◦C/m for water

ΔTf = 1.86 * 2.529

ΔTf = 4.70 C

5. the freezing point is therefore

0.00 ◦C - 4.70 ◦C = -4.70 ◦C

4 0

3 years ago

Read 2 more answers