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2 years ago

What is the similarity between teeth and eggshells?

2 answers:

6 0

Eggshells have a similar chemical composition to our tooth enamel, making them react similarly with other chemicals. This can help us understand what stains tooth enamel. When we brush an eggshell with fluoridated toothpaste, it strengthens the shell and protects it from acid, just like it does for our tooth enamel.

topjm [15]2 years ago

3 0

Explanation:

Because they share a similar composition, similar chemicals affect their structures in positive or negative ways. For example, fluoride – a staple in many dental practices – strengthens both enamel and eggshells and helps protect them from acids. Acids weaken and break down both substances. Scientists find this particularly concerning given that the ocean is growing increasingly acidic. They fear this may weaken the eggs of some marine species and harm their chance of survival. Most dentists recommend limiting aggressively acidic foods and beverages such as soft drinks.

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PLEASE HELP!!!!!!

pogonyaev

Answer:

$m_{B}^{theoretical}=0.365gB$

$Y=87.1\%$

Explanation:

Hello there!

In this case, since the reaction (A->B) have an initial amount of pure 4-aminobenzoic acid, the first step to compute the theoretical yield is to solve the following stoichiometric setup:

$m_{B}^{theoretical}=0.303gA*\frac{1molA}{137.14gA}*\frac{1molB}{1molA}*\frac{165.19 gB}{1molB}\\\\ m_{B}^{theoretical}=0.365gB$

Whereas A stands for 4-aminobenzoic acid and B for the benzocaine. Moreover, we compute the percent yield by dividing the actual yield (0.318 g) by the theoretical one (0.365 g):

$Y=\frac{0.318g}{0.365g} *100\%\\\\Y=87.1\%$

Best regards!

4 0

2 years ago

Hydrazine, N2H4, is a corrosive liquid used in rocket and jet fuels. Ammonia, NH3, is a gas that dissolves in water to form a so

Lyrx [107]

OK in the case of hydrazine 14 grams of nitrogen combine with 2 gram of hydrogen and with ammonia 14 grams combine with 3 grams of hydrogen.

Ratio 2:3

5 0

3 years ago

A container holds 100.0 mL of nitrogen at 21° C and a pressure of 736 mm Hg. What will be its volume if the temperature increase

devlian [24]

Answer:

V₂ = 104.76 mL

Explanation:

Given data:

Initial volume = 100.0 mL

Initial temperature = 21°C (21 + 273.15 K = 294.15 K)

Final temperature = 35°C (35 + 273.15 K = 308.15 k)

Final volume = ?

Solution:

Charles Law:

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ =100.0 mL × 308.15 K / 294.15 K

V₂ = 30815 mL.K /294.15 K

V₂ = 104.76 mL

5 0

2 years ago

The radiator in a car is filled with a solution of 60% antifreeze and 40% water. The manufacturer of the antifreeze suggests tha

lakkis [162]

Answer:

0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.

Explanation:

Volume of the radiator = 3.6 L

Percentage of antifreeze = 60%

Total volume of anti freeze in radiator = 60% of 3.6 L :

$\frac{60}{100}\times 3.6 = 2.16 L$

Percentage of water= 40%

Given,optimal cooling of the engine is obtained with only 50% antifreeze.

So, now we want to reduce the percentage of antifreeze from 60% to 50 %

Volume of coolant removed = x

Volume of water added = x

Volume of anti freeze removed = 60% of(x) = 0.6x

Volume of antifreeze left in radiator =50% of 3.6 L = $\frac{50}{100}\times 3.6=1.8 L$

1.8 Liter is the desired volume of the antifreeze.

Total volume - Removed volume = desired volume

2.16 L - 60% of( x) = 1.8 L

$2.16 L-0.6x=1.8 L$

$x=\frac{2.16 l- 1.8 L}{0.6}=0.6 L$

0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.

4 0

3 years ago

An object starts at position 12 on a horizontal line with a reference point of o. What is the position of the object if it moves

eimsori [14]

Answer:

-2

Explanation:

Consider object is starting 12 units right from the reference point which is 0.

Assign the right direction positive sign.

when object is moving 14 units on left direction the position of object will be two units to the left side of reference point.

Assign the left direction negative sign position will be -2.

6 0

3 years ago