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3 years ago

What is a cholesterol level of 1.85 g/l in the standard units of mg/dl?

2 answers:

8 0

Cholesterol is a sterol, it is a type of lipid molecule, and is biosynthesized by all animal cells. Total cholesterol level lesser than 200 milligrams per deciliter (mg/dL) is considered desirable for an adult. The standard unit used for any substance is gram per liter. However sometimes the concentration can also be expressed in terms of mg per deciliter. In this context 1 gram per liter is equal to 100 milligram per deciliter.

Thus 1.85 gram per liter is equal to 185 mg per deciliter. So 1.85 g/L is equivalent to 185mg per deciliter.

JulijaS [17]3 years ago

8 0

Answer: The cholesterol level is 185 mg/dL

Explanation:

We are given:

Cholesterol level = 1.85 g/L

To convert this level into milligrams per deciliters, we use the conversion factor:

1 g = 1000 mg

1 L = 10 dL

Converting the above speed into miles per hour, we get:

$\Rightarrow (\frac{1.85g}{L})\times (\frac{1000mg}{1g})\times (\frac{1L}{10dL})\\\\\Rightarrow 185mg/dL$

Hence, the cholesterol level is 185 mg/dL

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Calculate the mass of butane needed to produce 97.4 g of carbon dioxide. Express your answer to three significant figures and in

Vsevolod [243]

Answer:

32.1 g

Explanation:

Step 1: Write the balanced combustion reaction

C₄H₁₀ + 6.5 O₂ ⇒ 4 CO₂ + 5 H₂O

Step 2: Calculate the moles corresponding to 97.4 g of CO₂

The molar mass of CO₂ is 44.01 g/mol.

97.4 g × 1 mol/44.01 g = 2.21 mol

Step 3: Calculate the moles of butane that produced 2.21 moles of carbon dioxide

The molar ratio of C₄H₁₀ to CO₂ is 1:4. The moles of C₄H₁₀ required are 1/4 × 2.21 mol = 0.553 mol

Step 4: Calculate the mass corresponding to 0.553 moles of C₄H₁₀

The molar mass of C₄H₁₀ is 58.12 g/mol.

0.553 mol × 58.12 g/mol = 32.1 g

4 0

3 years ago

C. Calculate the number of moles in 62g of CO2

worty [1.4K]

Answer:

32÷5

I'm just tryna get points I'm sorry

goodluck tho❤

4 0

3 years ago

Read 2 more answers

What is the difference b/w thermodynamic cycle and mechanical cycle

const2013 [10]

In a mechanical cycle, mechanical energy (mostly the the rotation) is used to get the desired result. The form of energy remains the same.
In a thermodynamic cycle heat is converted to mechanical energy. That is to say there is conversion of energy.

4 0

4 years ago

Somebody please help me!!

mariarad [96]

Answer:

The given isotope is Br 80

If 35 protons present then its atomic number is 35

Mass number is 80 so its meutrons are 45 because proton+ neutron = mass number

Its electron are 36 because of negative one charge

Explanation:

4 0

3 years ago

Electrophilic addition of hbr to 3-methyl-2-hexene creates an asymmetric center at c-2. what is the product of this reaction?

AleksandrR [38]

This reaction would give rise to two products.

2-bromo-3-methylhexane, and
3-bromo-3-methylhexane.

However, 2-bromo-3-methylhexane would be more common than 3-bromo-3-methylhexane among the products.

The hydrogen atom in a hydrogen bromide molecule carries a partial positive charge. It is attracted to the double bond region with a high electron density. The hydrogen-bromine bond breaks when HBr gets too close to a double bond to produces a proton $\text{H}^{+}$ and a bromide ion $\text{Br}^{-}$ .

The proton would attack the double bond to produce a carbocation. It could attach itself to either the second or the third carbon atom.

$\phantom{\text{H}_3\text{C}-\text{CH}-\;}+\\\text{H}_3\text{C}-\text{CH}-\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3\\\phantom{\text{H}_3\text{C}-\;\;}|\phantom{\text{CH}-}\;}|\\\phantom{\text{H}_3\text{C}-\;} \text{H}\phantom{\text{CH}}\;\;}\text{CH}_3$
$\phantom{\text{H}_3\text{C}-\text{CH}-}\;\;\text{H}\\\phantom{\text{H}_3\text{C}-\;}+\phantom{\text{H}-\;}|\\\text{H}_3\text{C}-\text{CH}-\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3\\\phantom{\text{H}_3\text{C}-\;\;\;}\phantom{\text{CH}-}\;}|\\\phantom{\text{H}_3\text{C}-\;} \phantom{\text{H}}\phantom{\text{CH}}\;\;}\text{CH}_3$

Carbocations are unstable and might decompose over time. The first carbocation is more stable than the second for having three alkyl groups- i.e., straight carbon chains- attached to the center of the positive charge. Alkyl groups have stabilizing positive induction effect on positively-charged carbon. The second carbocation has only two, and is therefore not as stable. The first carbocation thus has the greatest chance to react with a bromide ion to produce a stable halocarbon.

Bromide ions are negatively charged. They attach themselves to carbocations at the center of positive charge. Adding a bromide ion to the first carbocation would produce 3-bromo-3-methylhexane whereas adding to the second produces 2-bromo-3-methylhexane.

The most likely product of this reaction is therefore 3-bromo-3-methylhexane.

6 0

3 years ago