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3 years ago

The formula of water, H2O suggests which of the following?

Chemistry

1 answer:

o-na [289]3 years ago

6 0

Answer: A. There are two hydrogen atoms and one oxygen atom per water molecule.

Explanation:

The chemical formular for water "H2O", means it contain 2 hydrogen atoms and one oxygen atom per molecule.

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If an electrostatically charged object is placed near other objects, which of the following will occur? Select all that apply. I

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(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g

Anarel [89]

Answer:

Part A

The volume of the gaseous product is $V = 787L$

Part B

The volume of the the engine’s gaseous exhaust is $V_e = 2178 \ L$

Explanation:

Part A

From the question we are told that

The temperature is $T = 350^oC = 350 +273 =623K$

The pressure is $P = 735 \ torr = \frac{735}{760} = 0.967\ atm$

The of $C_8 H_{18} = 100.0g$

The chemical equation for this combustion is

$2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}$

The number of moles of $C_8 H_{18}$ that reacted is mathematically represented as

$n = \frac{mass \ of \ C_8H_{18} }{Molar \ mass \ of C_8H_{18} }$

The molar mass of $C_8 H_{18}$ is constant value which is

$M = 114.23 \ g/mole$

So $n = \frac{100 }{114.23} }$

$n = 0.8754 \ moles$

The gaseous product in the reaction is $CO_2_{(g)}$ and water vapour

Now from the reaction

2 moles of $C_8 H_{18}$ will react with 25 moles of $O_2$ to give (16 + 18) moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

1 mole of $C_8 H_{18}$ will react with 12.5 moles of $O_2$ to give 17 moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

This implies that

0.8754 moles of $C_8 H_{18}$ will react with (12.5 * 0.8754 ) moles of $O_2$ to give (17 * 0.8754) of $CO_2_{(g)}$ and $H_2 O_{(g)}$

So the no of moles of gaseous product is

$N_g = 17 * 0.8754$

$N_g = 14.88 \ moles$

From the ideal gas law

$PV = N_gRT$

making V the subject

$V = \frac{N_gRT}{P}$

Where R is the gas constant with a value $R = 0.08206 \ L\cdot atm /K \cdot mole$

Substituting values

$V = \frac{14.88* 0.08206 *623}{0.967}$

$V = 787L$

Part B

From the reaction the number of moles of oxygen that reacted is

$N_o = 0.8754 * 12.5$

$N_o = 10.94 \ moles$

The volume is

$V_o = \frac{10.94 * 0.08206 *623}{0.967}$

$V_o = 579 \ L$

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

$V_e = V_o * \frac{0.79}{0.21}$

Substituting values

$V_e = 579 * \frac{0.79}{0.21}$

$V_e = 2178 \ L$

3 0

3 years ago

orbital cellulitis 5462 children included for study, 1347 (24%) were prescribed corticosteroids within 2 days of admission

son4ous [18]

It's unable to identify a decrease in LOS linked to corticosteroid exposure during hospitalization for ocular cellulitis in this database search. After two days of hospitalization, operational episodes and the prescription of corticosteroids were related to admission to the PICU.

Within two days of admission, 1347 (24%) of the 5462 children who were included in the research received a corticosteroid prescription. In analyses that controlled for age, the existence of meningitis, abscess, or visual problems, as well as the surgical episode and PICU admission within 2 days, corticosteroid prescription was not linked with LOS (e = 1.01, 95% confidence interval [CI]: 0.97-1.06). Among patients with a primary diagnosis of orbital cellulitis, corticosteroid exposure was linked to surgical events after two days of hospitalization (odds ratio = 2.05, 95% CI: 1.29-3.27) and 30-day readmission (odds ratio = 2.40, 95% CI: 1.52-3.78). Prospective, randomized control trials are required prior to the widespread usage of corticosteroids.

Learn more about orbital cellulitis here:-

brainly.com/question/15572218

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4 0

2 years ago