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3 years ago

Conduction transfers thermal energy by____.

Chemistry

2 answers:

diamong [38]3 years ago

6 0

Answer:

Explanation:

Conduction works by energy passing through touching objects

elixir [45]3 years ago

3 0

Moving energy in between two different objects in contact with each other

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A 2.684-g sample of zinc oxide was reduced by hydrogen gas, resulting in 2. 156 g of pure zinc metal. Determine the empirical fo

Afina-wow [57]

The empirical formula of the initial zinc oxide is ZnO.

<h3>What is Empirical Formula?</h3>

The empirical formula of a compound represents the ratios of elements in a compound but not the actual numbers or arrangement of the atoms.

It is the lowest whole number ratio of the element in the compound.

<h3>How to find out the empirical formula?</h3>

Find out the given masses and molar masses of the elements

The molar mass of Zn = 65 gmol⁻¹

Given the mass of Zn = 2.156 g

The molar mass of Oxygen = 16 gmol⁻¹

The mass of Oxygen = Mass of a sample of zinc oxide - the mass of zinc metal

= (2.684 - 2.156) g

= 0.528 g

Find the number of moles of the elements in the compound

The number of moles is given by

$n = \frac{m}{M}$

where m = given mass and

M = Molar mass

Number of moles of Zinc = $\frac{2.156}{65}$ = 0.033 moles

Number of moles of Oxygen = $\frac{0.528}{16}$ = 0.033 moles

Find the simplest ratios of the elements in the compound. To find the ratios simply divide the number of moles by the lowest number of moles obtained.

Here, the number of moles is the same for both elements. Hence, the simplest ratio for Zn:O is 1:1.

Therefore, the empirical formula of zinc oxide is ZnO.

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1 year ago

RIGHT ANWSER WILL BE MARKED BRAINLIEST

dybincka [34]

Answer:

11460

Explanation:

4 0

3 years ago

Part 1. Determine the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm. Show your work.

zaharov [31]

If a sample of gas is a 0.622-gram, volume of 2.4 L at 287 K and 0.850 atm. Then the molar mass of the gas is 7.18 g/mol

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates to the macroscopic properties of ideal gases.

An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Given :

V = 2.4 L = 0.0024

P = 86126.25 Pa

T = 287 K

m = 0.622

R = 8.314

The ideal gas equation is given below.

n = PV/RT

n = 86126.25 x 0.0024 / 8.314 x 287

n = 0.622 / molar mass (n = Avogardos number)

Molar mass = 7.18 g

Hence, the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm is 7.18 g

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1 year ago

A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?

Nastasia [14]

Answer:

At -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

where $P_{1}$ and $P_{2}$ are initial and final pressure respectively.

$V_{1}$ and $V_{2}$ are initial and final volume respectively.

$T_{1}$ and $T_{2}$ are initial and final temperature in kelvin scale respectively.

Here $P_{1}=150.0kPa$ , $V_{1}=1.75L$ , $T_{1}=(273-23)K=250K$ , $P_{2}=210.0kPa$ and $V_{2}=1.30L$

Hence $T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}$

$\Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}$

$\Rightarrow T_{2}=260K$

$\Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}$

So at -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

5 0

3 years ago

If hydrogen were used as a fuel, it could be burned according to the following reaction: H2(g)+1/2O2(g)→H2O(g). Use average bond

serious [3.7K]

The deltaHrxn = -243 kJ/mol the deltaHrxn of CH4(methane) = -802 kJ/mol

The fuel that yields more energy per mole is METHANE. The negative sign merely signifies the release of energy. Thus, 802 kJ/mol is greater than 243 kJ/mol.

The fuel that yields more energy per gram is HYDROGEN. Here is the computation:
deltaHrxn = (-243 kJ/mol)(1 mol/2.016 g H2) <span>= -120.535714286 kJ/g or -121 kJ/g

</span>deltaHrxn of CH4(methane) = (-802 kJ/mol)(1 mol/16.04 g)
<span>= -50 kJ/g
</span>
As discussed the negative sign serves as the symbol of released energy. Thus, 121 is greater than 50.

7 0

3 years ago