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3 years ago

What determines your blood cholesterol level?

Chemistry

2 answers:

andre [41]3 years ago

4 0

Saturated fat
Trans fat
Total fat
Excess calories
Dietary cholesterol

Romashka-Z-Leto [24]3 years ago

4 0

The type of food you eat.

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A sample of ideal gas is in a sealed container. The pressure of the gas is 485 torr , and the temperature is 40 ∘C . If the temp

Fantom [35]

Answer:

537.68 torr.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and V are constant, and have different values of P and T:

(P₁T₂) = (P₂T₁).

P₁ = 485 torr, T₁ = 40°C + 273 = 313 K,

P₂ = ??? torr, T₂ = 74°C + 273 = 347 K.

∴ P₂ = (P₁T₂)/(P₁) = (485 torr)(347 K)/(313 K) = 537.68 torr.

3 0

3 years ago

I need the answer to 452 of arogon

yaroslaw [1]

I'm confused... is there more info?

3 0

3 years ago

Which is the term for how vegetation influences precipitation?

Scilla [17]

The answer to this question is A- evaporation

3 0

3 years ago

An acid with the general formula RCOOH is used to make a 0.10 M solution in water. The pH of this acid solution is measured as 3

Sati [7]

Answer:

0.159 \%

Explanation:

The acid will dissociate according to the reaction shown below:-

$RCOOH + H_2O\rightleftharpoons RCOO^- + H_3O^+$

Given that, pH=3.8

The concentration of can be determined from the expression fo pH as:-

pH = - log $[H^+]$

3.8 = - log $[H^+]$

$[H^+]$ = $1.59\times 10^{-4}\ M$

The initial concentration of RCOOH was 0.10 M, then the percent dissociation was- calculated as shown below:-

$\% \ dissociation=\frac{1.59\times 10^{-4}}{0.10}\times 100=0.159\ \%$

3 0

3 years ago

Researchers used a combustion method to analyze a compound used as an antiknock additive in gasoline. A 9.394 mg sample of the c

LuckyWell [14K]

Answer:

The percent composition of the compound is 90.5 % C and 9.5 % H

Explanation:

Step 1: Data given

Mass of compound = 9.394 mg

Mass of CO2 yielded = 31.154 mg

Mass of H2O yielded = 7.977 mg

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: Calculate moles CO2

moles of CO2 = (0.031154 g / 44.01 g/mol) = 7.08 * 10^-4 mol CO2

Step 3: Calculate moles C

moles of C = moles of CO2 * (1 mol C / 1 mol CO2)

moles of C = 7.08 * 10^-4 mol

Step 4: Calculate moles H2O

moles of H2O = (0.007977 g / 18.02 g/mol) = 4.43 * 10^-4 mol H2O

Step 5: Calculate moles of H

moles of H = moles of H2O * (2 mol H / 1 mol H2O)

moles of H = 4.43* 10^-4 *2 = 8.86 * 10^-4 mol H

Step 6: Calculate mass of C

mass C = moles C * molar mass C

mass C = 7.08 * 10^-4 mol*12.01 g/mol

mass C = 0.0085 grams

Step 7: Calculate mass of H

mass H = moles H * molar mass H

mass H = 8.86 * 10^-4 mol*1.01 g/mol

mass H = 0.000894 grams

Step 8: Calculate total mass of compound =

0.0085 grams + 0.000894 grams = 0.009394 grams = 9.394 mg

Step 9: Calculate the percent composition:

% C = (8.50 mg / 9.394 mg) x 100 = 90.5%

% H = (0.894 mg / 9.394 mg) x 100 = 9.5%

The percent composition of the compound is 90.5 % C and 9.5 % H

6 0

3 years ago