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2 years ago

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For a school event 1/6 of the athletic field is reversed for the fifth -grade classes the reserved part of the field is divided

equally Among the 4 fifth grade classes in the school what fraction of the whole athletic field is reserved for each fifth class

1 answer:

SVETLANKA909090 [29]2 years ago

3 0

Answer:

$\frac{1}{24}$

Explanation:

Given:

For a school event, 1/6 of the athletic field is reserved for the fifth -grade classes and the reserved part of the field is divided equally among the 4 fifth grade classes in the school.

To find: fraction of the whole athletic field reserved for each fifth class

Solution:

Fraction of the whole athletic field reserved for four fifth classes = $\frac{1}{6}$

So, fraction of the whole athletic field reserved for each fifth class = $\frac{1}{4}(\frac{1}{6})=\frac{1}{24}$

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Sulfur dioxide has a vapor pressure of 462.7 mm Hg at –21.0 °C and a vapor pressure of 140.5 mm Hg at –44.0 °C. What is the enth

stealth61 [152]

Answer : The value of $\Delta H_{vap}$ is 28.97 kJ/mol

Explanation :

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = vapor pressure at temperature $-21.0^oC$ = 462.7 mmHg

$P_2$ = vapor pressure at temperature $-44.0^oC$ = 140.5 mmHg

$\Delta H_{vap}$ = Enthalpy of vaporization = ?

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = $-21.0^oC=[-21.0+273]K=252K$

$T_2$ = final temperature = $45^oC=[-41.0+273]K=232K$

Putting values in above equation, we get:

$\ln(\frac{140.5mmHg}{462.7mmHg})=\frac{\Delta H_{vap}}{8.314J/mol.K}[\frac{1}{252}-\frac{1}{232}]\\\\\Delta H_{vap}=28966.6J/mol=28.97kJ/mol$

Therefore, the value of $\Delta H_{vap}$ is 28.97 kJ/mol

4 0

3 years ago

Be sure to answer all parts. The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calcu

pochemuha

Answer : The volume of $CO_2$ will be, 514.11 ml

Explanation :

The balanced chemical reaction will be,

$HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2$

First we have to calculate the mass of $HCO_3^-$ in tablet.

$\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% \times 3.79g=\frac{32.5}{100}\times 3.79g=1.23175g$

Now we have to calculate the moles of $HCO_3^-$ .

Molar mass of $HCO_3^-$ = 1 + 12 + 3(16) = 61 g/mole

$\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles$

Now we have to calculate the moles of $CO_2$ .

From the balanced chemical reaction, we conclude that

As, 1 mole of $HCO_3^-$ react to give 1 mole of $CO_2$

So, 0.0202 mole of $HCO_3^-$ react to give 0.0202 mole of $CO_2$

The moles of $CO_2$ = 0.0202 mole

Now we have to calculate the volume of $CO_2$ by using ideal gas equation.

$PV=nRT$

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = $37^oC=273+37=310K$

n = number of moles of gas = 0.0202 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get :

$(1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)$

$V=0.51411L=514.11ml$

Therefore, the volume of $CO_2$ will be, 514.11 ml

6 0

2 years ago

PLS HELP ME I HAVE TILL 12

MariettaO [177]

Answer:

alright bud lets see hmm.... the answer is a. 90.5kpa

Explanation:

92.3 kPa - 1.82 kPa = 90.5 kPa

keep up your hope also corrected me if a am wrong in anyway! :)

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2 years ago

Which gas will effuse at the rate closest At a particular pressure and temperature, nitrogen gas effuses at the rate of 79mLs. U

Contact [7]

Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

$R\propto \sqrt{\frac{1}{M}}$

or,

$(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}$ ..........(1)

where,

$R_1$ = rate of effusion of nitrogen gas = $79mL/s$

$R_2$ = rate of effusion of sulfur dioxide gas = ?

$M_1$ = molar mass of nitrogen gas = 28 g/mole

$M_2$ = molar mass of sulfur dioxide gas = 64 g/mole

Now put all the given values in the above formula 1, we get:

$(\frac{79mL/s}{R_2})=\sqrt{\frac{64g/mole}{28g/mole}}$

$R_2=52mL/s$

Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.

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2 years ago

If a fluorine atom and nitrogen atom combine to form a compound, what type of bond will they form

dmitriy555 [2]

Answer:

I guess covalent bond is formed

4 0

3 years ago