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3 years ago

Describe how to separate a mixture of sugar and water.

2 answers:

4 0

Evaporation can separate sugar and water the water will evaporate leaving you a sugar a dried up sugar scrub at the bottom of the pan how to evaporate you can put the sugar water on a pan and boil it well water evaporates leaving it to boil for a long time low back route the water to leave you with only sugar

Nonamiya [84]3 years ago

3 0

Let the water evaporate and you'll be left with the sugar, or boil the water and collect the steam. the sugar will be left in the container you boiled the water in and you'll still have the water from the steam.

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If gatorade g series, endurance formula, contains 14 grams of carbohydrate per 8 ounces, how many ounces should a basketball pla

Mashcka [7]

Based on recommended amount of carbohydrate, a basketball player should consume about 17 - 34 ounces of gatorade g series during the hour-long game.

<h3>How many ounces of endurance formula gatorade g series, endurance formula should a basketball player consume during an hour-long game if it contains 14 grams of carbohydrate per 8 ounces?</h3>

Carbohydrates are food substances metabolized easily by the body to produce energy.

Given that the recommended amount of carbohydrate to consume to maintain performance is 30–60 g/h.

Also 14 grams of carbohydrate found in 8 ounces of the drink.

30 g of carbohydrate will be present in 30 × 8/14 = 17.1 ounces of gatorade g series

60 g of carbohydrate will be present in 60 × 8/14 =34.3 ounces of gatorade g series.

Therefore, a basketball player should consume about 17 - 34 ounces of gatorade g series during the hour-long game.

Learn more about carbohydrates at: brainly.com/question/797978

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2 years ago

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.