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Alla [95]
3 years ago
7

Nathan placed 1 cup of sand (S), potting soil, (P), and a mixture of sand and soil (M) into separate containers. In each of the

containers he placed a thermometer so that the bulb was 2.5 cm below the surface. He place the 3 containers under identical heat lamps for an hour. The original temperature of each container was 15C. After heating the jars in three separate trials, the temperatures of the containers were:
Chemistry
1 answer:
lina2011 [118]3 years ago
3 0

Answer:

Sand: 28°C, 27°C, 26°C

Potting soil: 33°C, 29°C, 31°C

Mixture: 29°C, 29°C, 22.5°C.

Explanation:

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inn [45]

Answer:

fatty acids

Explanation:

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A 56 g sample of peanut oil has a volume of 60.9 cm³. Calculate the density of the peanut oil
aleksandr82 [10.1K]
Density = mass / volume. So you'd write it out as D = 56g / 60.9cm^3, giving you 0.9195 g/cm^3.
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I NEED HELP PLEASE! :)<br><br> Describe a limiting reactant in your own words.
koban [17]

Answer:

See explanation

Explanation:

A limiting reactant is the reactant in a reaction that is done reacting first, because there are less moles of it than are needed for a full reaction with the other compound or compounds.

Hope this helps!

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3 years ago
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A 35.0 ml sample of 0.225 m hbr was titrated with 42.3 ml of koh. What is the concentration of the koh?
Lemur [1.5K]

Answer:

The concentration of KOH is 0.186 M

Explanation:

First things first, we need too write out the balanced equation between HBr and KOH.

This is given as;

KOH (aq) + HBr (aq) → KBr (aq) + H2O (l)

From the reaction above, we can tell that it takes 1 mole of KOH to react with 1 mole of HBr.

We use the acid base formular in calculating unknown concentrations. This is given as;

\frac{CaVa}{CbVb}  = \frac{na}{nb}

where;

Ca = Concentration of acid

Va = Volume of acid

Cb = Concentration of base

Vb = Volume of base

na = Number of moles of acid

nb = Number of moles of base

KOH is the base and HBr is acid.

Hence;

Ca = 0.225

Va = 35

Cb = ?

Vb = 42.3

na = 1

nb = 1

Making Cb subject of formular we have;

Cb = \frac{CaVaNb}{VbNa}

Cb = (0.225 * 35 * 1) / (42.3 * 1)

Cb = 0.186 M

5 0
3 years ago
How many molecules are in 7.34 g of Na2CO3
kenny6666 [7]

Answer:

The answer is 105.98844.

<h3>Explanation: </h3>

We assume you are converting between grams Na2CO3 and mole. You can view more details on each measurement unit: molecular weight of Na2CO3 or mol This compound is also known as Sodium Carbonate.

7 0
3 years ago
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