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2 years ago

A mixture consists of sand and an aqueous salt

1 answer:

4 0

The right answer for the question that is being asked and shown above is that: "(1) Evaporate the water, then filter out the salt." A mixture consists of sand and an aqueous salt solution. The procedure that can be used to separate the sand, salt, and water from each other is to evaporate the water, then filter out the salt.

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Just Lemons Lemonade Recipe Equation:

zalisa [80]

Answer:

Explanation:

Hello!

Complete text:

Honors Stoichiometry Activity WorksheetInstructions:

Activity Two: Just Lemons, Inc. Production

Here's a one-batch sample of Just Lemons lemonade production. Determine the percent yield and amount of leftover ingredients for lemonade production and place your answers in the data chart.

Hint: Complete stoichiometry calculations for each ingredient to determine the theoretical yield. Complete a limiting reactant-to-excess reactant calculation for both excess ingredients.

Water 946.36 g

Sugar 196.86 g

Lemon Juice 193.37 g

Lemonade 2050.25g

Leftover Ingredients?

Just Lemons Lemonade Recipe Equation:

2 water + sugar + lemon juice = 4 lemonade

Mole conversion factors:

1 mole of water = 1 cup = 236.59 g

1 mole of sugar = 1 cup = 225 g

1 mole of lemon juice = 1 cup = 257.83 g

1 mole of lemonade = 1 cup = 719.42 g

You have the information on the ingredients used to produce one batch of lemonade and the amount of lemonade produced. To determine which ingredients be leftovers, you have to determine first, which one is the limiting reactant, i.e. the ingredient that will be used up first.

According to the recipe, to make 4 moles of lemonade, you use 2 moles of water, one mole of sugar and one mole of lemon juice, expressed in grams:

2 water + sugar + lemon juice = 4 lemonade

2*(236.59) + 225g + 257.83g = 4*(719.42)g

473.18g + 225g + 257.83g = 2877.68g

So for every 2877.68g of lemonade made, they use 473.18g of water, 225g of sugar, and 257.83g of lemon juice.

You know that they made a batch of 2050.25g, so to detect the limiting reactant, first, you have to calculate, in theory, how much of each ingredient you need to make the given amount of lemonade:

Use cross multiplication

Water:

2877.68g lemonade → 473.18g water

2050.25g lemonade → X= (2050.25*473.18)/2877.68= 337.12g water

Following the recipe, to elaborate 2050.25g of lemonade, you need to use 337.12g of water.

Sugar:

2877.68g lemonade → 225g sugar

2050.25g lemonade → X= (2050.25*225)/2877.68= 160.30g sugar.

To elaborate 2050.25f of lemonade you need to use 160.30g of sugar.

Lemon juice:

2877.68g lemonade → 257.83g lemon juice

2050.25g lemonade → X= (2050.25*257.83)/2877.68= 183.69g lemon juice.

To elaborate 2050.25f of lemonade you need to use 183.69g lemon juice.

Available ingredients vs. theoretical yields for 2050.25g of lemonade:

Water 946.36 g → 337.12g

Sugar 196.86 g → 160.30g

Lemon Juice 193.37 g → 183.69g

The lemon juice will be the first ingredient to be used up, there will be a surplus of water and sugar.

I hope this helps!

7 0

2 years ago

A rigid, 28-L steam cooker is arranged with a pressure relief valve set to release vapor and maintain the pressure once the pres

MaRussiya [10]

Answer:

$\Delta S_{source}>-1.204\frac{kJ}{K}$

Explanation:

Hello!

In this case, given the initial conditions, we first use the 10-% quality to compute the initial entropy:

$s_1=s_{f,175kPa}+q*s_{fg,175kPa}\\\\s_1=1.4850\frac{kJ}{kg*K} +0.1*5.6865\frac{kJ}{kg*K}=2.0537\frac{kJ}{kg*K}$

Now the entropy at the final state given the new 40-% quality:

$s_2=s_{f,150kPa}+q*s_{fg,150kPa}\\\\s_2=1.4337\frac{kJ}{kg*K} +0.4*5.7894\frac{kJ}{kg*K}=3.7495\frac{kJ}{kg*K}$

Next step is to compute the mass of steam given the specific volume of steam at 175 kPa and the 10% quality:

$m_1=\frac{0.028m^3}{(0.001057+0.1*1.002643)\frac{m^3}{kg} } =0.274kg\\\\m_2=\frac{0.028m^3}{(0.001053+0.4*1.158347)\frac{m^3}{kg} } =0.0603kg$

Then, we can write the entropy balance:

$\Delta S_{source}+\frac{Q}{T_1} -\frac{Q}{T_2} +s_2m_2-s_1m_1-s_{fg}(m_2-m_1)>0$

Whereas sfg stands for the entropy of the leaving steam to hold the pressure at 150 kPa and must be greater than 0; thus we plug in:

Which is such minimum entropy change of the heat-supplying source.

Best regards!

3 0

2 years ago

According to reference table adv-10, which reaction will take place spontaneously?

olga_2 [115]

Missing table!! write the elements with the first letter of the symbol with Upper Caps letters!!!

http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm

Ni2+ +Pb(s) → Ni(s) + Pb2+
The potential of the oxidation of Pb(s) --> Pb2+(aq) is 0.126 V
The potential of the reduction go Ni2+(aq) --> Ni(s) is -0.25 V

Add the two together and the potential for the reaction is -0.124 V (NO SPONTANEOUS THE SIGN IS NEGATIVE)

au3+ + al(s) → au(s) + al3+Au3+(aq) -> Au(s) +1.5 VAl -> Al3+ +1.66VV= 3.16 (SPONTANEOUS THE SIGN OF THE PONTENTIAL IS POSITIVE)Sr2+ + Sn(s) → Sr(s) + Sn2+

Sr2+(aq) + 2 e– Sr(s) V= -2.89V
Sn -> Sn2+ V= 0.14 V
V= -2.75 V (no spontaneous)

Fe2+ + Cu(s) → Fe(s) + Cu2+
Fe2+(aq) + 2 e– Fe(s) V= -0.44 V
Cu -> C2+ V = - 0.337V

V= - 0.777V (no spontaneous)

5 0

2 years ago

in collecting the precipitate, why would it be inappropriate to heat the reacted mixture and evaporate off the water?

djverab [1.8K]

In collecting the precipitate, it is inappropriate to heat the reacted mixture and evaporate off the water because it is possible that the mixture contains other substances that precipitates as well when the mixture is being heated so you will not be able to collect what you want.

6 0

2 years ago

Electroplating is a way to coat a complex metal object with a very thin (and hence inexpensive) layer of a precious metal, such

8_murik_8 [283]

Answer:

0.0164 g

Explanation:

Let's consider the reduction of silver (I) to silver that occurs in the cathode during the electroplating.

Ag⁺(aq) + 1 e⁻ → Ag(s)

We can establish the following relations.

1 A = 1 C/s
The charge of 1 mole of electrons is 96,468 C (Faraday's constant)
1 mole of Ag(s) is deposited when 1 mole of electrons circulate.
The molar mass of silver is 107.87 g/mol

The mass of silver deposited when a current of 0.770 A circulates during 19.0 seconds is:

$19.0s \times \frac{0.770c}{s} \times \frac{1mole^{-} }{96,468C} \times \frac{1molAg}{1mole^{-}} \times \frac{107.87g}{1molAg} = 0.0164 g$

5 0

2 years ago