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3 years ago

help plz really need your help if u know answer if u don't DO NOT ANSWER this is not the first time I have posted this question

Chemistry

1 answer:

morpeh [17]3 years ago

6 0

T eg(1) : 97.5°C, T eg(2): 98.5°C, T eg(3): 99.2°C

∆T water(1): -2.5°C, ∆T water(2): -1.5°C, ∆T water(3): -0.8°C

∆T metal(1): 77.5°C, ∆T metal(2): 80.5°C, ∆T metal(3): 80.2°C.

ft= (m1 cp1 t1 + m2 cp2 t2 + .... + mn cpn tn) / (m1 cp1 + m2 cp2 + .... + mn cpn) (1)

where,

1000g = 1kg

ft(t eg)= final mixed temperature (°C)

m = mass of substance (kg)

cp = specific heat of substance (J/kg°C)

t = temperature of substance (°C)

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If 11g of a gas occupies 5.6dm'3at s.t.p., calculate it's vapour density (1.0mol of a gas occupies 22.4dm'3at s.t.p.).

11111nata11111 [884]

Answer:

${ \boxed{ \bf{vapour \: density = 2 \times molecular \: mass}}} \\{ \tt{ PV= (\frac{m}{ m_{r}}) RT}} \\ { \tt{3 \times 5.6 = \frac{11}{m _{r}} \times 0.0831 \times 273}} \\ { \tt{m _{r} = 14.85 \: g}} \\ \\ { \bf{vapour \: density = 2 \times m _{r}}} \\ = 2 \times 14.85 \\ = 29.7 \: { \tt{g {dm}^{ - 3} }}$

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3 years ago

A gas has a pressure of 5.7 atm at 100.0°C. What is its pressure at20.0°C (Assume volume is unchanged)

son4ous [18]

Answer:

$\large \boxed{\text{4.5 atm}}$

Explanation:

The volume and amount of gas are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

$\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}$

Data:

p₁ =5.7 atm; T₁ = 100.0 °C

p₂ = ?; T₂ = 20.0 °C

Calculations:

1. Convert the temperatures to kelvins

T₁ = (100.0 + 273.15) K = 373.15

T₂ = (20.0 + 273.15) K = 293.15

2. Calculate the new pressure

$\begin{array}{rcl}\dfrac{5.7}{373.15} & = & \dfrac{p_{2}}{293.15}\\\\0.0153 & = & \dfrac{p_{2}}{293.15}\\\\0.0153\times 293.15 &=&p_{2}\\p_{2} & = & \textbf{4.5 atm}\end{array}\\\text{The new pressure will be $\large \boxed{\textbf{4.5 atm}}$}$

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3 years ago

A cube has an edge length of 9 cm .

Angelina_Jolie [31]

Answer:

9^3 i think so like 279

Explanation:

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3 years ago

Specific gravity of a vegetable oil is 0.85. What is its density?

elena-s [515]

I think its density is 850kg/m^3

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3 years ago

A sample of gaseous neon atoms at atmospheric pressure and 0 °c contains 2.69 * 1022 atoms per liter. the atomic radius of neon

omeli [17]

Explanation

Radius of neon atom : 69 pm = $69\times 10^{-12} m$

Volume occupied by the one atom: $\frac{4}{3}\pi r^3$

$=\frac{4}{3}\times 3.14\times(69\times 10^{-12} m)^3=1.37\times 10^{-30} m^3$

given that $2.69\times 10^{22}$ atoms are present in 1L

1 L = 0.001 $m^3$

The volume occupied by the $2.69\times 10^{22}$ neon atoms

$2.69\times 10^{22}\times 1.37\times 10^{-30} m^3=3.68\times 10^{-8} m^3$

Fraction of volume occupied by the neon atom:

$=\frac{3.68\times 10^{-8} m^3}{0.001 m^3}=3.68\times 10^{-11} m^3$

$3.68\times 10^{-11} m^3$

The fraction of of volume occupied by the neon atom is very less than the 1 L which indicates the presence of large amount of empty space between the atoms of the gas.

3 0

3 years ago

help plz really need your help if u know answer if u don't DO NOT ANSWER this is not the first time I have posted this question​

help plz really need your help if u know answer if u don't DO NOT ANSWER this is not the first time I have posted this question