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olya-2409 [2.1K]
2 years ago
9

I have made a thermometer which measures temperature by the compressing and expanding of gas in a piston. I have measured that a

t 1000 C the volume of the piston is 30 L. What is the temperature outside if the piston has a volume of 5 L? What would be appropriate clothing for the weather? I
this is about boyle’s and charles law :)))
Chemistry
1 answer:
Scrat [10]2 years ago
6 0

Answer:

The temperature is 298.5 K, which corresponds to 0.50°C.

A jacket would be appropriate clothing for this weather.

Explanation: Hope this helps:)

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A 0.539 g sample of a metal, M, reacts completely with sulfuric acid according to the reaction M(s)+H2SO4(aq)⟶MSO4(aq)+H2(g) A v
Charra [1.4K]

Answer:

The molar mass of the metal is 54.9 g/mol.

Explanation:

When we work with gases collected over water, the total pressure (atmospheric pressure) is equal to the sum of the vapor pressure of water and the pressure of the gas.

Patm = Pwater + PH₂

PH₂ = Patm - Pwater = 1.0079 bar - 0.03167 bar = 0.9762 bar

The pressure of H₂ is:

0.9762bar.\frac{1atm}{1.013bar} =0.9637atm

The absolute temperature is:

K = °C + 273 = 25°C + 273 = 298 K

We can calculate the moles of H₂ using the ideal gas equation.

P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{0.9637atm \times 0.249L }{(0.08206atm.L/mol.K)\times298K} =9.81 \times 10^{-3} mol

Let's consider the following balanced equation.

M(s) + H₂SO₄(aq) ⟶ MSO₄(aq) + H₂(g)

The molar ratio of M:H₂ is 1:1. So, 9.81  × 10⁻³ moles of M reacted. The molar mass of the metal is:

\frac{0.539g}{9.81 \times 10^{-3} mol} =54.9g/mol

4 0
4 years ago
Assume that the density and heat of combustion of E85 can be obtained by using 85 % of the values for ethanol and 15 % of the va
Alika [10]

Answer:

87.4 J

Explanation:

The density of the gasoline is 0.70 g/mL, and the density of the ethanol is 0.79 g/mL. The heat combustions (the heat released in a combustion reaction) are 5,400 kJ/mol for gasoline, and 1,370 kJ/mol for ethanol.

For 3.5 L of E85, the volumes of gasoline and ethanol are:

Vgasoline = 0.15 * 3.5 = 0.525 L = 5.25x10⁻⁴ mL

Vethanol= 0.85 * 3.5 = 2.975 L = 2.975x10⁻³ mL

The mass of gasoline and ethanol presented in that sample of E85 is the volume multiplied by the density:

mgasoline = 5.25x10⁻⁴ * 0.70 = 3.675x10⁻⁴ g

methanol = 2.975x10⁻³ * 0.79 = 2.35025x10⁻³ g

The number of moles for each substance is it mass divided by its molar mass. The molar masses are 114 g/mol for gasoline, and 46 g/mol for ethanol:

ngasoline = 3.675x10⁻⁴/114 = 3.224x10⁻⁶ mol

nethanol = 2.35025x10⁻³ /46 = 5.109x10⁻⁵ mol

The energy released is the heat combustion multiplied by the number of moles, so:

Egasoline = 5,400 * 3.224x10⁻⁶ = 0.0174 kJ = 17.4 J

Eethanol = 1,370 * 5.109x10⁻⁵ = 0.07 kJ = 70 J

So, the energy released by the E85 is the sum of the energy released by ethanol and gasoline:

The energy released by E85 = 87.4 J

8 0
3 years ago
The amount of calcium hydroxide needed to react completely with 100.0g of nitric acid
Pani-rosa [81]

The amount of Calcium hydroxide : = 58.719 g

<h3>Further explanation</h3>

Given

100 g Nitric Acid-HNO₃

Required

The amount of Calcium hydroxide

Solution

Reaction(balanced) :

Ca(OH)₂ (s) + 2HNO₃ (aq) → Ca(NO₃)₂ (aq) + 2H₂O (l)

mol of Nitric acid (MW 63 g/mol) :

mol = mass : MW

mol = 100 : 63

mol = 1.587

From the equation, mol ratio of Ca(OH)₂ : HNO₃ = 1 : 2, so mol Ca(OH)₂ :

=1/2 x mol HNO₃

= 1/2 x 1.587

=0.7935

Mass of Ca(OH)₂ (MW=74 g/mol) :

= mol x MW

= 0.7935 x 74

= 58.719 g

5 0
3 years ago
Examine the typical street scene on the right. Draw and fill in the table. In column A, name five activities shown in which ther
valkas [14]

so a cat and a mouse walk in to a store

5 0
4 years ago
Gold is a very soft metal that can be hammered into extremely thin sheets known as gold leaf. If a 1.78 g piece of gold is hamme
Solnce55 [7]

Answer:

2.0489\times 10^{-6} cm is the average thickness of the sheet.

Explanation:

Mass of gold ,m= 1.78 g

Volume of the gold = V

Density of the gold = D = 19.32g/cm^3

D=\frac{m}{V}=19.32g/cm^3=\frac{1.78 g}{V}

V = 0.09213 cm^3

Area of the hammered gold sheet,A = 48.4 ft^2=44,965.052 cm^2

Thickness of the hammered gold = h

(1 ft^2=929.03 cm^2)

Volume = Area × thickness

V= A\times h

0.09213 cm^3=44,965.052 cm^2\times h

h=2.0489\times 10^{-6} cm

2.0489\times 10^{-6} cm is the average thickness of the sheet.

3 0
4 years ago
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