All categories

2 years ago

Is this statement true or false? Gymnosperms reproduce using seeds, but angiosperms do not.

Chemistry

1 answer:

spayn [35]2 years ago

3 0

Answer:

FALSE!

Explanation:

Gymnosperms do use seeds but are exposed like the pine cones of pines. Angiosperms still have seeds, however, they flower or fruit.

You might be interested in

1. What is the difference between accidental discoveries and scientific investigations?

Alenkasestr [34]

Answer:

the discovery will always need to be tested more where if you're investigating you should already be in a more controlled environment

4 0

3 years ago

1 cholo 2 floro 4 nitro + Na^+

N76 [4]

Answer:

Incomplete Q.

GOOD LUCK FOR THE FUTURE! :)

4 0

2 years ago

Read 2 more answers

What is the distance from earth to Neptune in standard notation and scientific notation.

ira [324]

Answer:

Scientific Notation: 4.45963 x 109 km (29.811 A.U.) Scientific Notation: 4.53687 x 109 km (30.327 A.U.) By Comparison: If you weigh 100 pounds on Earth, you would weigh 110 pounds on Neptune.

4 0

3 years ago

Read 2 more answers

Hey man forgets that he said his coffee cup on the top of his car he starts to drive in the coffee cup rolls off the car onto th

Lyrx [107]

It would move due to it not being the same speed and in the same direction so it is acted upon by an unbalanced force.

8 0

3 years ago

In the summer of 2016, the city of Columbus Dublin Road Water Treatment plan exceeded the regulatory level of nitrate, which is

Verizon [17]

Answer:

a) $10 mg NO_3^-/L$

b) $1.61*10^{-4}mol NO_3^-/L$

c) $2.26 mg N/L$

Carbon: $C=26.64 \frac{mg C}{L}$

Explanation:

Nitrate

First of all, is important to know that:

$1 ppm=1 mg/L$

a) 10 ppm of nitrate ( $NO_3^-$ ) is equal to $10 mg NO_3^-/L$

b) The molecular weight of nitrate is $62 g NO_3^-/mol$

$10 mg NO_3^-/L=0.01 g NO_3^-/L$

$\frac{0.01 g NO_3^-/L}{62 g NO_3^-/mol}=1.61*10^{-4} mol NO_3^-/L$

c) Nitrate has 14 mg of N per 62 mg of NO3

$10 mg NO_3^-/L*\frac{14 mg N}{62 mg NO_3^-}=2.26 mg N/L$

Carbon

Carbonate has 12 mg of C per 60 mg of $CO_3^{-2}$

Bicarbonate has 12 mg of C per 61 mg of $HCO_3^{-}$

$C=24 \frac{mg CO_3^{-2}}{L}*\frac{12 mg C}{60 mg CO_3^{-2}}+111 \frac{mg HCO_3^{-}}{L}*\frac{12 mg C}{61 mg HCO_3^{-}}$

$C=26.64 \frac{mg C}{L}$

8 0

3 years ago