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77julia77 [94]
3 years ago
13

25.0 mL of 2.00M solution are diluted to 85.0 ml. What is the new concentration?

Biology
1 answer:
k0ka [10]3 years ago
7 0

Explanation:

hi I don't think it is so bad for the advice of a good one and I was actually getting ready for u too many problems that u guys are so a diff one of my favorite things to catch up on but it was quite the best i mean turkish series are the best i mean seties are u guys were in my friends named had I don't think it was a joke anymore ik when I see you guys have it on ur highlight of the time I feel like I don't didn't say u have a great day and I was actually getting ready for the first time I was in yr was like a dog and that's what u mean by our convos n everything fe

hope this helped!:) if it did pls brainlist

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Which of these is unique to flowering plants?haploid gametophytes,an embryo surrounded by nutritive tissue,double fertilization,
fiasKO [112]

Answer:

double fertilization

Explanation:

Flowering plants or angiosperms are seed-producing plants with the ability to produce reproductive organs-flowers and fruits with seed in it (unlike gimnosperms which contain naked seed). Another distinctive feature of angiosperms is their reduced gametophytes. This feature most likely reduces the time between pollination and fertilization. Fertilization in flowerin plants is double, meaning that two sperm cells fertilize ovule cells(egg cell and central nuclei cell): one forms diploid zygote which will develop in embryo, while other form triploid cell which will develop into endosperm (provides nutrition for the embryo).

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4 years ago
Yeast cells are recovered from a fermentation broth by using a tubular centrifuge. Sixty percent of the cells are recovered at a
Serga [27]

Answer:

\\ \omega = 5033.22 \ rpm

Explanation:

Given that:

R_1 = 0.6\\\\R_2 = 0.95\\\\\omega_1 = 4000 \ rpm\\\\\omega_2 = ???\\\\Q_1 = 12\ l/min\\\\Q_2 =  12\ l/min

For a continuous centrifuge ; solid recovery R ∝ \frac{\omega^2}{Q}

where;

ω = angular velocity of the centrifuge in rpm

Q = flow rate

SO;

R_1 \alpha \frac{4000^2}{12}

R_2 \alpha  \frac{\omega^2}{12}

\frac{R_1}{R_2} = \frac{4000^2}{12}  ÷  \frac{\omega^2 }{12}

\frac{R_1}{R_2} = \frac{4000^2}{12} * \frac{12}{\omega^2}

\frac{R_1}{R_2} = \frac{4000^2}{\omega^2}\\\\\frac{0.6}{0.95} = \frac{4000^2}{\omega^2}

{\omega^2} = \frac{4000^2*0.95}{0.6}\\\\\\{\omega} = \sqrt{ \frac{4000^2*0.95}{0.6}}\\\\\\ \omega = 5033.22 \ rpm

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