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3 years ago

What is the purpose of strawberry seeds

Chemistry

1 answer:

Neporo4naja [7]3 years ago

7 0

Answer:

The goal there is for the fruit to not get eaten, so that the seed can rely on the fruit's nutrients to support its growth.” Presumably, the strawberry went for evolutionary option number one – attract something to spread the seeds

Explanation:

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A 30.2 mL aliquot of a 30.0 wt% aqueous KOH solution is diluted to 1.20 L to produce a 0.173 M KOH solution. Calculate the densi

alexandr1967 [171]

Answer:

The density of solution is 1.283 g/mL.

Explanation:

Molarity of the KOH before dilution = $M_1$

Volume of the solution before dilution = $V_1=30.2 mL$

Molarity of the KOH after dilution = $M_2=0.173 M$

Volume of the solution after dilution = $V_2=1.20 L=1200 mL$

$M_1V_1=M_2V_2$

$M_1=\frac{M_2V_2}{V_1}=\frac{0.173 M\times 1200 mL}{30.2 mL}$

$M_1=6.8742 M$

$Molarity=\frac{moles}{Volume (L)}$

$V_1=30.2 mL=0.0302 L$ (1 mL = 0.001 L)

$M_1=\frac{n}{V_1}$

$n=M_1\times V_1=6.8742 M\times 0.0302 L=0.2076 mol$

Mass of 0.2076 moles of KOH:

0.2076 mol × 56 g/mol = 11.6256 g

Mass of KOH is solution = 11.6265 g

Mass of the solution = M

Mass percentage of solution = 30.0% of KOH

$30.0\%=\frac{11.6265 g}{M}\times 100$

M = 38.755 g

Density of the solution , d= $\frac{M}{V_1}$

$d=\frac{38.755 g}{30.2 mL}=1.283 g/mL$

The density of solution is 1.283 g/mL.

5 0

3 years ago

Assume that the reaction of aqueous hydrobromic acid solution and potassium hydroxide base undergoes a complete neutralization r

Tamiku [17]

Answer:

$HBr + KOH → KBr + H _{2} O$

$1000 \: ml \: contains \: 0.685 \: moles \\ 55.4 \: ml \: contains \: ( \frac{55.4 \times 0.685}{1000} ) \\ = 0.038 \:moles \\ 1 \: mole \: of \: hydrobromic \: acid \: produces \: 1 \: mole \: of \: water \\ 0.038 \: moles \: produce \: (0.038 \times 1) \\ = 0.038 \: moles \\ 1 \: mole \: of \: water \: weighs \: 18 \: g \\ 0.038 \: moles \: weighs \: (0.038 \times 18) \: g \\ = 0.684 \: g$

0.042 M

1.4

5 0

2 years ago

Copper(II) sulfate pentahydrate, CuSO4 ·5 H2O, (molar mass 250 g/mol) can be dehydrated by repeated heating in a crucible. Which

prohojiy [21]

Answer:

The water lost is 36% of the total mass of the hydrate

Explanation:

Step 1: Data given

Molar mass of CuSO4*5H2O = 250 g/mol

Molar mass of CuSO4 = 160 g/mol

Step 2: Calculate mass of water lost

Mass of water lost = 250 - 160 = 90 grams

Step 3: Calculate % water

% water = (mass water / total mass of hydrate)*100 %

% water = (90 grams / 250 grams )*100% = 36 %

We can control this by the following equation

The hydrate has 5 moles of H2O

5*18. = 90 grams

(90/250)*100% = 36%

(160/250)*100% = 64 %

The water lost is 36% of the total mass of the hydrate

8 0

2 years ago

What are the products of the following equation?

Artyom0805 [142]

The products are on the right side of the equation. For this one it would be 2AlPO4 + 3CaSO4

5 0

2 years ago

Adding energy to a solid bar of gold may result in which of the following outcomes

____ [38]

An increase in motion and less attraction between particles

4 0

2 years ago