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Sladkaya [172]
3 years ago
6

Suppose you want to find out the temperature at which different kinds of liquid freeze. What do you need in order to perform the

experiment? What is the variable in this experiment?
Chemistry
1 answer:
Nana76 [90]3 years ago
3 0

In order to perform the experiment, in identifying the temperature of different kinds of liquid freeze—the needed equipment is the thermometer. It is because the thermometer is an equipment that helps in determining a temperature. The variable in the experiment would be the temperature as this is not consistent for the effects or outcome will likely change.

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Help please asap. chem sucks
luda_lava [24]

I'm assuming that C is carbon.

4.590 \: mol \: c \times  \frac{12.01 \:g \: c}{1 \: mol \: c}

55.1259 g of C

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3 years ago
In Period 2, as the elements are considered from left to
Elza [17]

Answer:

it will option B ,hope it helps

5 0
3 years ago
Explain how cells use digested food in your own words
natima [27]

Answer:

The digestive system uses mechanical and chemical methods to break food down into nutrient molecules that can be absorbed into the blood. ... Some animals use intracellular digestion, where food is taken into cells by phagocytosis with digestive enzymes being secreted into the phagocytic vesicles.

Explanation:

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3 years ago
A 5.00g of X, the product of organic synthesis is obtained in a 1.0 dm3 aqueous solution. Calculate the mass of X that can be ex
Anestetic [448]

Answer:

mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g

Explanation:

The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula

K = concentration of X in ether/concentration of X in water

Partition coefficient, K(X) between ethoxy ethane and water = 40

Concentration of X in ether = mass(g)/volume(dm³)

Mass of X in ether = m g

Volume of ether = 50/1000 dm³ = 0.05 dm³

Concentration of X in ether = (m/0.05) g/dm³

Concentration of X in water = mass(g)/volume(dm³)

Mass of X in water left after extraction with ether = (5 - m) g

Volume of water = 1 dm³

Concentration of X in water = (5 - m/1) g/dm³

Using K = concentration of X in ether/concentration of X in water;

40 = (m/0.05)/(5 - m)

(m/0.05) = 40 × (5 - m)

(m/0.05) = 200 - 40m

m = 0.05 × (200 - 40m)

m = 10 - 2m

3m = 10

m = 10/3

m = 3.33 g of X

Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g

8 0
3 years ago
Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2
Setler79 [48]

Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol

7 0
3 years ago
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