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nekit [7.7K]
4 years ago
5

What are the five steps a flavor chemist needs to review before starting a flavor project

Chemistry
1 answer:
Katyanochek1 [597]4 years ago
5 0

Answer:

flavorists, work primarily in the food, beverage, pharmaceutical, cosmetic, and pet industries. They use scientific and analytical tools, along with creativity, to boost natural flavors and create new ones

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What can humans do to reduce climate change
yanalaym [24]

Answer:

Ten possibilities for staving off catastrophic climate change ... What can one person, or even one nation, do on their own to slow and reverse climate change? ... could reduce greenhouse gas emissions to safer levels—there are ... mode of transport that does not require anything other than human energy.

Explanation:

6 0
4 years ago
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BRAINLIEST PLEASE HELP!!!
Natalka [10]
Answer : first opinion and also last
Expiation : Note that these last two reactions, and 2H + 2H → 4He + γ, .Nuclear fusion is a reaction in which two nuclei are combined to form a larger nucleus. Nuclear fusion is a reaction in which two nuclei are combined, or fused, to form a larger nucleus. We know that all nuclei have less mass than the sum of the masses of the protons and neutrons that form them. The missing mass times c2 equals the binding energy of the nucleus—the greater the binding energy, the greater the missing mass.
5 0
3 years ago
When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.
sp2606 [1]

Answer:

Mass of CaCl₂ =  20 g

CaCO is presewnt in excess.

Mass of of CaCO₃ remain unreacted =  7.007 g

Explanation:

Given data:

Mass of calcium carbonate = 25 g

Mass of hydrochloric acid = 13.0 g

Mass of calcium chloride produced = ?

Chemical equation:

CaCO₃ + 2HCl  →  CaCl₂  + H₂O + CO₂

Number of moles of CaCO₃:

Number of moles of CaCO₃ = Mass /molar mass

Number of moles of CaCO₃= 25.0 g / 100.1 g/mol

Number of moles of CaCO₃ = 0.25 mol

Number of moles of HCl:

Number of moles of  HCl = Mass /molar mass

Number of moles of HCl = 13.0 g / 36.5 g/mol

Number of moles of HCl = 0.36 mol

Now we will compare the moles of CaCl₂ with HCl and CaCO₃ .

                  CaCO₃         :               CaCl₂

                    1                 :               1

                 0.25              :            0.25

                HCl                :                CaCl₂

                 2                   :                    1

                 0.36            :                  1/2 × 0.36 = 0.18 mol

The number of moles of CaCl₂ produced by HCl are less it will be limiting reactant.

Mass of CaCl₂ = moles × molar mass

Mass of CaCl₂ =0.18 mol × 110.98 g/mol

Mass of CaCl₂ =  20 g

The calcium carbonate is present in excess.

                HCl                :                CaCO₃

                 2                   :                    1

                 0.36            :                  1/2 × 0.36 = 0.18 mol

So, 0.18 moles react with 0.36 moles of HCl.

The moles of CaCO₃ remain unreacted = 0.25 -0.18

The moles of CaCO₃ remain unreacted = 0.07 mol

Mass of of CaCO₃ remain unreacted = Moles × molar mass

Mass of of CaCO₃ remain unreacted = 0.07 mol × 100.1 g/mol

Mass of of CaCO₃ remain unreacted =  7.007 g

7 0
3 years ago
Design a test to determine whether thorium-234 also emits particles. First, explain how Rutherford’s experiment measured positiv
liubo4ka [24]

The characteristics of the α and β particles allow to find  the design of an experiment to measure the ²³⁴Th particles is:

  • On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the beta particle emission from ²³⁴Th.
  • The neutrons cannot be detected in this experiment because they have no electrical charge.

In Rutherford's experiment, the positive particles directed to the gold film were measured on a phosphorescent screen that with each arriving particle a luminous point is seen.

The particles in this experiment are α particles that have two positive charge and two no charged is a helium nucleus.

The test that can be carried out is to place a small ours of Thorium in front of a phosphorescent screen and see if it has flashes, with the amount of them we can determine the amount of particle emitted per unit of time.

Thorium has several isotopes, with different rates and types of emission:

  • ²³²Th emits α particles, it is the most abundant 99.9%
  • ²³⁴Th emits β particles, exists in small traces.

In this case they indicate that the material used is ²³⁴Th, which emits β particles that are electrons, the detection of these particles is more difficult since it has one negative charge, it has much lower mass, but they can travel further than the particles α, therefore, for what type of isotope we have, we can start measuring at a small distance and increase the distance until the reading is constant. At this point all the particles that arrive are β, which correspond to ²³⁴Th.

Neutron detection is much more difficult since these particles have no charge and therefore do not interact with electrons and no flashing on the screen is varied.

In conclusion with the characteristics of the α and β particles we can find the design of an experiment to measure the ²³⁴Th particles is:

  • On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the β particle emission from ²³⁴Th.
  • The neutrons cannot be detected in this experiment because they have no electrical charge.

Learn more about radioactive emission here: brainly.com/question/15176980

7 0
3 years ago
If 10.62 mL of a standard 0.3330 M KOH solution reacts with 98.20 mL of CH3COOH solution, what is the molarity of the acid solut
Nikolay [14]

Answer:

0.036 M of CH_{3} COOH

Explanation:

It is an example of acid-base neutralization reaction.

KOH  + CH_{3} COOH  ----> CH_{3} COO^{-} K^{+}   +   H_{2}O

Base           Acid                           Salt                                    

When two component react then the number of moles of both the component should be same, therefore the number of moles and acids and bases should be the same in the following .

Molarity= \frac{\textrm{No. of Moles}}{\textrm{Volume of the Particular Solution}}

No.of moles= Molarity × Volume of the Particular Solution

Therefore,

M_{1}V_{1} =M_{2}V_{2}------------------------------(1)

where

M_{1}= Molarity of Acid

V_{1}= Volume of Acid

M_{2}= Molarity of Base

V_{2}= Volume of Base

M_{1}=0.3330 M

V_{1}=10.62 mL

V_{2}=98.2 mL

M_{2}=??(in M)

Plugging in Equation 1,

0.3330 × 10.62 =M_{2}  × 98.2  

M_{2}=\frac{0.3330*10.62}{98.2}

M_{2}=0.036 M

3 0
3 years ago
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