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Akimi4 [234]
2 years ago
7

Describe the sequence of adding ingredients to make the recipe for condensed tomato soup. To mix the ingredients, you have an in

dustrial-sized vat with a capacity of 800 L. You also have the ability to heat what is in the vat and/or stir the vat, both of which require energy and therefore, cost you money. Focus on the fact that the manufacturing process should be time efficient and cost efficient in terms of energy use. Be sure to refer to the solubility charts or graphs that you find for salt and sugar during your research. Finally, presume that the tomato powder has low solubility which is not temperature dependent. What changes could be made to optimize the manufacturing process?  i need to know the last sentence of this question 
Chemistry
1 answer:
mezya [45]2 years ago
3 0
The final sentence for this question would be: "Finally, presume that the tomato powder has low solubility which is not temperature dependent."
So to answer the question, 360 grams of NaCl can be dissolved in 1 L water. So, 2000 grams sugar can be dissolved in 1 L water then we can say that the solubility of salt is lesser in water as to sugar and both heightened by increasing the temperature. If we make a batch of 800 L we can add sugar, 1600 kg at 25 0c. We can add salt is 288 kg at 25 0c and the ingredient tomato is having low solubility.
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7. You are about to perform some intricate electrical studies on single skeletal muscle fibers from a gastronemius muscle. But f
Vilka [71]

Answer:

58.61 grams

Explanation:

Taking The molecular weight of NaCl = 58.44 grams/mole

<u>Determine how many grams of NaCl to prepare the bath solution </u>

first we will calculate the moles of NaCl that is contained in 6L of 170 mM of NaCI solution

= ( 6 * 170 ) / 1000

= 1020 / 1000 = 1.020 moles

next

determine how many grams of NaCl

= moles of NaCl * molar mass of NaCl

= 1.020 * 58.44

= 58.61 grams

4 0
2 years ago
Plasmid DNA and a gene of interest are cut with the enzyme PpuMI. Write a possible sequence of bases for the sticky end of the g
nasty-shy [4]

Answer:

5' RG        GWCCY 3'

3' YCCWG        GR 5'

Explanation:

The enzyme PpuMI is a restriction endonuclease enzyme, it has a specific recognition site where it cut the DNA. The source of the enzyme is from ​​an E. coli strain that carries the PpuMI gene from Pseudomonas putida (R. Morgan).

The enzyme PpuMI recognizes specific sequence with palindrome arrangement. It target the sequence 5' RGGWCCY 3'

target Sequence: 5' RGGWCCY 3'

                            3' YCCWGGR 5'

The enzyme cleavage point is at:

5' RG^GWCCY 3'

3' YCCWG^GR 5'

The product of the cleavage will give a sticky end Cleavage:

5' RG        GWCCY 3'

3' YCCWG        GR 5'

Note: R stands for purines (adenine and guanine). Y stands for pyrimidines (cytosine, thymine, and uracil). And W represents adenine or thymine.

5 0
2 years ago
How many grams of NH3 can be prepared from the synthesis of 77.3 grams of nitrogen and 14.2 grams of hydrogen gas?
lbvjy [14]

Answer:

80.41 g

Explanation:

Data Given:

Mass of Nitrogen (N₂) = 77.3 g

Mass of Hydrogen (H₂) = 14.2 g

many grams of NH₃ = ?

Solution:

First we look at the balanced synthesis reaction

              N₂   +    3 H₂  ------—> 2 NH₃

             1 mol      3 mol

As 1 mole of Nitrogen react with 3 mole of hydrogen

Convert moles to mass

molar mass of N₂ = 2(14) = 28 g/mol

molar mass of H₂ = 2(1) + 2 g/mol

Now

                     N₂             +           3 H₂        ------—>      2 NH₃

             1 mol (28 g/mol)     3 mol(2g/mol)

                    28 g                        6 g

28 grams of N₂ react with 6 g of H₂  

So

if 28 grams of N₂ produces 6 g of H₂  so how many grams of N₂ will react with 14.2 g of H₂.

Apply Unity Formula

                 28 g of N₂ ≅ 6 g of H₂

                 X g of N₂ ≅ 14.2 g of H₂

Do cross multiply

                X g of N₂ = 28 g x 14.2 g / 6 g

                X g of N₂ = 66.3 g

As we have given with 77. 3 g of N₂ but from this calculation we come to know that 66.3 g will react with 14.2 g of hydrogen and the remaining 10 g N₂ will be in excess

So, Hydrogen is limiting reactant in this reaction and the amount of NH₃ depends on the amount of hydrogen.

Now

To find mass of NH₃ we will do following calculation

Look at the reaction

As we Know

                     N₂             +           3 H₂        ------—>      2 NH₃

                                                   6 g                            2 mol

So, 6 g of hydrogen gives 2 moles of NH₃, then how many moles of NH₃ will be produce by 14.2 g

Apply Unity Formula

                 6 g of H₂ ≅ 2 mol of NH₃

                14.2 g of H₂ ≅ X mol of NH₃

Do cross multiply

               X mol of NH₃= 14.2 g x 2 mol / 6 g

                X mol of NH₃ = 4.73 mol

So, 14.2 g of hydrogen gives 4.73 moles of NH₃

Now

Convert moles of NH₃ to mass

Formula will be used

        mass in grams = no. of moles x molar mass . . . . . . (2)

Molar mass of  NH₃

Molar mass of  NH₃ = 14 + 3(1)

Molar mass of  NH₃ = 14 + 3 = 17 g/mol

Put values in equation 2

        mass in grams = 4.73 mole x 17 g/mol

        mass in grams =  80.41 g

mass of NH₃=  80.41 g

3 0
2 years ago
A 0.580 g sample of a compound containing only carbon and hydrogen contains 0.480 g of carbon and 0.100 g of hydrogen. At STP, 3
Sati [7]

Answer:

Molecular formula for the gas is: C₄H₁₀

Explanation:

Let's propose the Ideal Gases Law to determine the moles of gas, that contains 0.087 g

At STP → 1 atm and 273.15K

1 atm . 0.0336 L = n . 0.082 . 273.15 K

n = (1 atm . 0.0336 L) / (0.082 . 273.15 K)

n = 1.500 × 10⁻³ moles

Molar mass of gas = 0.087 g / 1.500 × 10⁻³ moles = 58 g/m

Now we propose rules of three:

If 0.580 g of gas has ____ 0.480 g of C _____ 0.100 g of C

58 g of gas (1mol) would have:

(58 g . 0.480) / 0.580 = 48 g of C  

(58 g . 0.100) / 0.580 = 10 g of H

 48 g of C / 12 g/mol = 4 mol

 10 g of H / 1g/mol = 10 moles

7 0
2 years ago
The reaction between nitrogen dioxide and carbon monoxide is NO2(g)+CO(g)→NO(g)+CO2(g)NO2(g)+CO(g)→NO(g)+CO2(g) The rate constan
Alex17521 [72]

Answer : The rate constant at 525 K is, 0.0606M^{-1}s^{-1}

Explanation :

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 701K = 2.57M^{-1}s^{-1}

K_2 = rate constant at 525K = ?

Ea = activation energy for the reaction = 1.5\times 10^2kJ/mol=1.5\times 10^5J/mol

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 701 K

T_2 = final temperature = 525 K

Now put all the given values in this formula, we get:

\log (\frac{K_2}{2.57M^{-1}s^{-1}})=\frac{1.5\times 10^5J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{701K}-\frac{1}{525K}]

K_2=0.0606M^{-1}s^{-1}

Therefore, the rate constant at 525 K is, 0.0606M^{-1}s^{-1}

8 0
2 years ago
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