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andrew11 [14]
3 years ago
13

Agar is used to culture bacteria and as a medium in electrophoresis. The agar used in this Petri dish is a by-product of an unli

kely organism. Agar is extracted from a(n)
Chemistry
1 answer:
Ymorist [56]3 years ago
4 0
Agar is extracted from an algae (agarophytes, <span>belong to the </span>Rhodophyta (red algae) phylum, <span>primarily from the </span>genera Gelidium<span> and </span><span>Gracilaria</span>).
Agar <span>or </span>agar-agar is a jelly-like substance. Agar is the mixture of two components, agaropectin (D-glucuronic acid and pyruvic acid) and <span>the polysaccharide agarose.</span>
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If 10.0g of oxygen reacts with 10.0g of hydrogen, how many grams of water can be
rusak2 [61]

Answer:

the answer is 90g

Explanation:

2g of H2 produce 18g of H2O/10.0g of H2 to produce x the answer is 90g

8 0
3 years ago
What mass of salt (nacl) should you add to 1.48 l of water in an ice cream maker to make a solution that freezes at -13.4 ∘c ? a
blagie [28]
Answer is: mass of salt is 311,15 g.
V(H₂O) = 1,48 l · 1000 ml/l = 1480 ml.
m(H₂O) = 1480 g = 1,48 kg.
d(solution) = 1,00 g/ml.
ΔT(solution) = 13,4°C = 13,4 K.
Kf = 1,86 K·kg/mol; cryoscopic constant of water
i(NaCl) = 2; Van 't Hoff factor.
ΔT(solution) = Kf · b · i.
b(NaCl) = 13,4 K ÷ (1,86 K·kg/mol · 2).
b(NaCl) = 3,6 mol/kg.
n(NaCl) = 3,6 mol · 1,48 kg= 5,328 mol.
m(NaCl) = 5,328 mol · 58,4 g/mol = 311,15 g.

5 0
2 years ago
Read 2 more answers
Is it possible to see the tiny particles that make up a grain of sand
77julia77 [94]
Technically speaking, yes you can. Using a microscope though.
6 0
3 years ago
Read 2 more answers
2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s
Aleks04 [339]

Answer:

In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

                       = 2 mL

<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

6 0
3 years ago
What mass of sodium bicarbonate do you start with
valentina_108 [34]

Answer:

84.007 g/mol

Explanation:

welcome

5 0
3 years ago
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