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3 years ago

A rock has a mass of 15 grams and a volume of 5 cm3 what is the density of the object

Chemistry

1 answer:

Zina [86]3 years ago

5 0

Answer:

The answer is

Explanation:

To find the density of a substance when given the mass and volume we use the formula

<h3> $density = \frac{mass}{volume}$ </h3>

From the question

mass = 15 g

volume of rock = 5 cm³

The density of the substance is

$density = \frac{15}{5}$

We have the final answer as

Hope this helps you

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What is the mass of 6.2 mil of K2CO3

Oksi-84 [34.3K]

Answer:

6.2 moles of K2CO3 can be converted to 856.8741 grams.

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3 years ago

Determine the volume of methane (ch 4 ) gas needed to react completely with 0.660 l of o 2 gas to form methanol (ch 3 oh).

Sloan [31]

Balance Chemical Equation for this reaction is,

2 CH₄ + O₂ → 2CH₃OH

According to this eq, 22.4 L (1 moles) of Oxygen requires 44.8 L (2 mole) CH₄ for complete reaction.
So, the volume of CH₄ required to consume 0.66 L of O₂ is calculated as,

22.4 L O₂ required to consume = 44.8 L CH₄
0.660 L O₂ will require = X L of CH₄

Solving for X,
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X = 1.320 L of CH₄

Result:
1.320 L of CH₄ <span>gas is needed to react completely with 0.660 L of O</span>₂<span> gas to form methanol (CH</span>₃OH<span>).</span>

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3 years ago

Electroplating is a way to coat a complex metal object with a very thin (and hence inexpensive) layer of a precious metal, such

8_murik_8 [283]

Answer:

0.0164 g

Explanation:

Let's consider the reduction of silver (I) to silver that occurs in the cathode during the electroplating.

Ag⁺(aq) + 1 e⁻ → Ag(s)

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1 A = 1 C/s
The charge of 1 mole of electrons is 96,468 C (Faraday's constant)
1 mole of Ag(s) is deposited when 1 mole of electrons circulate.
The molar mass of silver is 107.87 g/mol

The mass of silver deposited when a current of 0.770 A circulates during 19.0 seconds is:

$19.0s \times \frac{0.770c}{s} \times \frac{1mole^{-} }{96,468C} \times \frac{1molAg}{1mole^{-}} \times \frac{107.87g}{1molAg} = 0.0164 g$

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3 years ago

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Answer : The correct option is, +91 kJ/mole

Solution :

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$Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)$

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^0_{[Pb^{2+}/Pb]}=-0.13V$