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2 years ago

What is the definition of heat?

1 answer:

4 0

Heat is the transfer of kinetic energy from one medium or object to another, or from an energy source to a medium or object. Such energy transfer can occur in three ways: radiation, conduction, and convection.

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The last intermediate in citric acid cycle is:_________

GenaCL600 [577]

The last intermediate in citric acid cycle is Oxaloacetic acid.

<h3>What is Citric Acid Cycle?</h3>

Organic molecule HOC(CO2H)(CH2CO2H)2 is the chemical formula for citric acid. It is a weak organic acid that is colorless. Citrus fruits naturally contain it. It is a biochemical intermediary in the citric acid cycle, which is a component of all aerobic organisms' metabolism.

Every year, more than two million tons of citric acid are produced. It is frequently used as a flavoring, an acidifier, and a chelating agent.

Citrates, which include salts, esters, and the polyatomic anion present in solution, are derivatives of citric acid. Trisodium citrate is an example of the former; triethyl citrate is an example of an ester.

Learn more about citric acid with the help of the given link:

brainly.com/question/15582668

#SPJ4

7 0

4 months ago

a sample of an oxide of antimony (sb) contain 19.75 g of antimony combine with 6.5 g of oxygen . what is the simplest formula fo

frez [133]

Explanation:

The given data is as follows.

Mass of antimony = 19.75 g

Molar mass of Sb = 121.76 g/mol

Therefore, calculate number of moles of Sb as follows.

Moles of Sb = $\frac{mass}{\text{molar mass}}$

= $\frac{19.75 g}{121.76 g/mol}$

= 0.162 mol

Mass of oxygen given is 6.5 g and molar mass of oxygen is 16 g/mol. Hence, moles of oxygen will be calculated as follows.

Moles of oxygen = $\frac{mass}{\text{molar mass}}$

= $\frac{6.5 g}{16 g/mol}$

= 0.406 mol

Hence, ratio of moles of Sb and O will be as follows

Sb : O

$\frac{0.162}{0.162} : \frac{0.406}{0.162}$

1 : 2.5

We multiply both the ratio by 2 in order to get a whole number. Therefore, the ratio will be 2 : 5.

Thus, we can conclude that the empirical formula of the given oxide is $Sb_{2}O_{5}$ .

4 0

2 years ago

3. The doctor notices that Janet has long fingernails, and comments that fingernails grow at a rate (or

Viktor [21]

Answer:

7.93 × 10⁻¹³ km/s

Explanation:

Step 1: Given data

Rate of growth of the fingernails (r): 2.50 cm/year

Step 2: Convert "r" from cm/year to km/year

We will use the following conversion factors.

1 m = 100 cm

1 km = 1000 m

$\frac{2.50cm}{year} \times \frac{1m}{100cm} \times \frac{1km}{1000m} = 2.50 \times 10^{-5} km/year$

Step 3: Convert "r" from km/year to km/s

We will use the following conversion factors.

1 year = 365 day

1 day = 24 h

1 h = 60 min

1 min = 60 s

$\frac{2.50 \times 10^{-5} km}{year} \times \frac{1year}{365day} \times \frac{1day}{24h} \times \frac{1h}{60min} \times \frac{1min}{60s} = 7.93 \times 10^{-13} km/s$

2.50 cm/year is equal to 7.93 × 10⁻¹³ kilometers/second.

6 0

2 years ago

Caustic soda is 19.1 M NaOH and is diluted for household use. What is the household concentration if 10 mL of the concentrated s

olga nikolaevna [1]

Hello!

To calculate the household concentration of NaOH we need to use the dilution formula, clearing for M2, as you can see in the equation below:

$M1*V1=M2*V2 \\ \\ V2= \frac{M1*V1}{V2}$

Now, we input the values from the data we have onto this equation. M1=19,1 M; V1=10 mL; V2=400 mL, and solve the equation to get the result:

$V2= \frac{M1*V1}{V2}= \frac{19,1M*10mL}{400 mL}=0,48M$

So, the household concentration of NaOH will be 0,48 M

Have a nice day!

4 0

3 years ago

What volume, in milliliters, of calcium chloride 15.0 M stock solution would you use to

Kamila [148]

<u>Answer:</u> The volume of stock solution of calcium chloride required is 10 mL.

<u>Explanation:</u>

A solution consists of solute and solvent. A solute is defined as the component that is present in a smaller proportion while the solvent is defined as the component that is present in a larger proportion.

To calculate the amount of solute needed, the formula used is:

$C_1V_1=C_2V_2$ ....(1)