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3 years ago

The following reaction shows sodium carbonate reacting with calcium hydroxide.

Chemistry

1 answer:

velikii [3]3 years ago

5 0

Ans: 15.1 grams

Given reaction:

Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3

Mass of Na2CO3 = 20.0 g

Molar mass of Na2CO3 = 105.985 g/mol

# moles of Na2CO3 = 20/105.985 = 0.1887 moles

Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH

# moles of NaOH produced = 0.1887*2 = 0.3774 moles

Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol

Mass of NaOH produced = 0.3774*39.996 = 15.09 grams

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Determine which choice is an example of an endothermic process.

klemol [59]

Answer:

D. Baking bread

Explanation:

In this process, energy is absorbed and in an endothermic process energy is absorbed too.

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3 years ago

What agricultural is practiced in or near San Diego? Choose a food that is grown or produced locally. Describe the food,Where an

liraira [26]

Answer:

San Diego boasts top crops in nursery, flowers, avocados, tomatoes, citrus, chickens & eggs, shrooms, succulents, strawberries, coffee & cannabis. Most of San Diego County farms are small 1-9 acre proprieties, and that awards San Diego County more certified organic growers than any other county in the nation.

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2 years ago

List the following aqueous solutions in order of decreasing freezing point: 0.040 m glycerin (C3H8O3), 0.020 m potassium bromide

gogolik [260]

Answer:

Solutions from highest to lowest freezing point:

0.040 m glycerin = 0.020 m potassium bromide > 0.030 m phenol

Explanation:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f$ = Depression in freezing point

i = van'T Hoff fcator

$K_f$ = Molal depression constant of solvent

m = molality of the solution

Higher the value of depression in freezing point at lower will be freezing temperature the solution.

1. 0.040 m glycerin

Molal depression constant of water = $K_f=1.86^oC/m$

i = 1 ( organic molecule)

m = 0.040 m

$\Delta T_{f,1}=1\times\times 1.86^oC/m\times 0.040 m$

$\Delta T_{f,1}=0.0744^oC$

2. 0.020 m potassium bromide

Molal depression constant of water = $K_f=1.86^oC/m$

i = 2 (ionic)

m = 0.020 m

$\Delta T_{f,2}=2\times\times 1.86^oC/m\times 0.020 m$

$\Delta T_{f,2}=0.0744^oC$

3. 0.030 m phenol

Molal depression constant of water = $K_f=1.86^oC/m$

i = 1 (organic)

m = 0.030 m

$\Delta T_{f,3}=1\times\times 1.86^oC/m\times 0.030 m$

$\Delta T_{f,3}=0.0558^oC$

$0.0744^oC=0.0744^oC > 0.0558^oC$

$\Delta T_{f,1}=\Delta T_{f,2}>\Delta T_{f,3}$

Solutions from highest to lowest freezing point:

0.040 m glycerin = 0.020 m potassium bromide > 0.030 m phenol

5 0

3 years ago

2h2o2 (l) → 2h2o(g) + o2(g) how many grams of hydrogen peroxide are needed to produce 90.0 g of water?

skelet666 [1.2K]

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3 years ago

How many moles of gold, a U, are in 3.60×10^-5 G of gold?

ch4aika [34]

Answer:

See Below.

Explanation:

This is a conversion problem:

Using the molar mass of Gold (given in a periodic table) which is 196.97g/mol

you have

$(3.60*10^-5 g) *\frac{1moles}{196.97g}$

you always arrange the equation in a way to cancel whatever you don't want (grams) and leave what you do want (moles). Here grams cancel (top and bottom), so you're left with:

$\frac{(3.60*10^-5 g) *(1)}{196.97} = 1.83*10^-7$ moles of Gold

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4 years ago