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3 years ago

What are the uses of carbon dioxide

Chemistry

1 answer:

suter [353]3 years ago

7 0

Answer:

Could you give brainliest btw here you go

Explanation:

Uses of Carbon in daily life

It makes up for 18% of the human body. Sugar, glucose, proteins etc are all made of it. ...

Carbon in its diamond form is used in jewellery. ...

Amorphous carbon is used to make inks and paints. ...

Graphite is used as the lead in your pencils. ...

One of the most important uses is carbon dating.

a refrigerant, in fire extinguishers,

for inflating life rafts and life jackets,

blasting coal,

foaming rubber and plastics,

promoting the growth of plants in greenhouses,

immobilizing animals before slaughter, and in carbonated beverages.

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The Nutrition Facts label for crackers states that one serving contains 19 g of carbohydrate, 4 g of fat, and 2 g of protein. Wh

sp2606 [1]

Answer: The amount of kilocalories contained in the given serving of crackes is 0.120 kCal

Explanation:

As per the USDA:

Carbohydrates provide 4 calories per gram, protein provides 4 calories per gram, and fat provides 9 calories per gram

We are given:

Mass of fat in the crackers = 4 g

Mass of carbohydrates in the crackers = 19 g

Mass of protein in the crackers = 2 g

Conversion factor used: 1 kCal = 1000 Cal

Applying unitary method:

For Fat:

1 gram of fat provides 9 calories

So, 4 gram of fat will provide = $\frac{9}{1}\times 4=36Cal=0.036kCal$

For Carbohydrates:

1 gram of carbohydrates provides 4 calories

So, 19 gram of carbohydrates will provide = $\frac{4}{1}\times 19=76Cal=0.076kCal$

For Proteins:

1 gram of proteins provides 4 calories

So, 2 gram of fat will provide = $\frac{4}{1}\times 2=8Cal=0.008kCal$

Total kilocalories per serving = [0.036 + 0.076 + 0.008] kCal = 0.120 kCal

Hence, the amount of kilocalories contained in the given serving of crackes is 0.120 kCal

8 0

3 years ago

In which layer(s) would a solid cube with 6 cm sides and mass of 270 g float? Explain.

DIA [1.3K]

The buoyancy of an object is dictated by its density. So let us calculate for density, where:density = mass / volume
Calculate the volume first of a solid cube:volume = (6 cm)^3 = 216 cm^3 = 216 mL
Therefore density is:density = 270 g / 216 mLdensity = 1.25 g / mL
Therefore this object will float in the layer in which the density is more than 1.25 g / mL.

6 0

3 years ago

The protein lysozyme unfolds at a transition temperature of 75.5°C, and the standard enthalpy of transition is 509 kJ mol-1. Cal

spin [16.1K]

Answer:

0.4774 KJ/K.mol

Explanation:

We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii

Now,

C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol

Now, for the first process, ΔS_i is given as;

ΔS_i = C_p,m × In(T2/T1)

We are given;

T1 = 25°C = 25 + 273.15K = 298.15 K

T2 = 75.5°C = 75.5 + 273.15 K=348.65 K

Thus;

ΔS_i = C_p,m × In(348.65/298.15)

Now, for the third process, ΔS_iii is given as;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)

Thus;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)

Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;

In(298.15/348.65) can also be written as;

In(348.65/298.15)^(-1) or

- In(348.65/298.15)

Thus;

ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]

Now, let's add ΔS_iii to ΔS_i to get;

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]

First 2 terms will cancel out to give;

ΔS_i + ΔS_iii = -6.28In(348.65/298.15)

ΔS_i + ΔS_iii = -0.9826 KJ/K.mol

Now,for process ii;

ΔS_ii = standard enthalpy of transition/Transition Temperature

Thus;

ΔS_ii = (509 KJ/K.mol)/348.65

ΔS_ii = 1.46 KJ/K.mol

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii = -0.9826 + 1.46 = 0.4774 KJ/K.mol

5 0

3 years ago

Which of the following is an important intermediate in the mechanism of the reaction in the box?

Dennis_Churaev [7]

Answer:

Explanation:

3 0

3 years ago

Which of the following ions is formed when a base is dissolved in a solution?

Aneli [31]

Answer:

$\huge\boxed{\sf OH-}$

Explanation:

According to Arrhenius concept of acid and base:

"When a base in a solution, produces/yields OH- (Hydroxide) ions."

So, when a base is dissolved in a solution, it produces OH- ions.

For example:

NaOH ⇄ Na⁺ + OH⁻ (So, it is a base)

$\rule[225]{225}{2}$

Hope this helped!

3 0

2 years ago