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Contact [7]
3 years ago
14

4.85 x 10^-4 tons to cups

Chemistry
1 answer:
fiasKO [112]3 years ago
7 0

Answer:

2.08753919851 cups

0.464942810185139 cups

2.08753919851 cups

Explanation:

i looked it up

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Which of the following is a synthesis reaction?
nadezda [96]

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What is the estimates range of conduct disorder in boys?
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3 years ago
4. How many grams of ammonium carbonate are needed to decompose in order to produce
Thepotemich [5.8K]

Answer:

14.23g of (NH4)2CO3

Explanation:

We'll begin by writing the balanced equation for the reaction.

(NH4)2CO3 –> (NH4)2O + CO2

Next,, we shall determine the mass of (NH4)2CO3 that decomposed and the mass of CO2 produced from the balanced equation. This is illustrated below:

Molar mass of (NH4)2CO3 = 2[14+(4x1)] + 12 + (16x3)

= 2[14 +4] + 12 + 48

= 2[18] + 60 = 96g/mol

Mass of (NH4)2CO3 from the balanced equation = 1 x 96 = 96g

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of CO2 from the balanced equation = 1 x 44 = 44g.

Summary:

From the balanced equation above,

96g of (NH4)2CO3 decomposed to produce 44g of CO2.

Finally, we can determine the mass of (NH4)2CO3 that decomposed to produce 6.52g of CO2 as follow:

From the balanced equation above,

96g of (NH4)2CO3 decomposed to produce 44g of CO2.

Therefore, Xg of (NH4)2CO3 will decompose to produce 6.52g of CO2 i.e

Xg of (NH4)2CO3 = (96 x 6.52)/44

Xg of (NH4)2CO3 = 14.23g

Therefore, 14.23g of (NH4)2CO3 is needed to produce 6.52g of CO2.

4 0
3 years ago
Winter is coming and it's time to make sure you have the right amount of antifreeze
ololo11 [35]

The molality of the solution = 17.93 m

<h3>Further explanation</h3>

Given

6.00 L water with 6.00 L of  ethylene glycol(ρ=1.1132 g/cm³= 1.1132 kg/L)

Required

The molality

Solution

molality = mol of solute/ 1 kg solvent

mol of solute = mol of ethylene glycol

  • mass of ethylene glycol :

= volume x density

= 6 L x 1.1132 kg/L

= 6.6792 kg

= 6679.2 g

  • mol of ethylene glycol (MW=62.07 g/mol)

=mass : MW

=6679.2 : 62.07

=107.608

  • mass of water

6 L water = 6 kg water(ρ= 1 kg/L)

  • molality

\tt =\dfrac{107.608}{6}=17.93~m

5 0
2 years ago
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