If you have 8.5 moles of water (H2O), how many grams of water do you have?

The specific heat of copper is 0.385 j/(g · °c). if 34.2 g of copper, initially at 24.0°c, absorbs 4.689 kj, what will be the fi

julia-pushkina [17]

We know the relation between heat and temperature follows the formula: Q = Ce * m * (Tf-Ti) , where Q is heat transfer, Ce is specific heat, m is mass, Tf is final temperature and Ti is initial temperature. Note that heat value is given in kj so we need to change to joules, then as heat is absorbed the final temperature will increase. 4,689 = 0.385 * 34.2 * (Tf-24) Tf = 4,689 / (0.385 * 34.2) + 24 =380.12ÂşC

7 0

4 years ago

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The Ka of an acid is 1.3 x 10–7. Based on the Ka and its relationship with Kw, what is the value of Kb?

Solnce55 [7]

Ka x Kb = Kw Kw = 1 x 10⁻¹⁴

(1.3 x 10⁻⁷) x Kb = 1 x 10⁻¹⁴

Kb = ( 1 x 10⁻¹⁴ ) / ( 1.3 x 10⁻⁷)

Kb = 7.7 x 10⁻⁸

Answer D

hope this helps!

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3 years ago

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Completion

Snowcat [4.5K]

Answer:

Compress
Fixed
Melts
Melting Point
Freezing Point
High
Crystalline
Lattice
Unit cell
Amorphous solids

Explanation:

Solids tend to be dense and difficult to compress.

They do not flow or take the shape of their containers, like liquids do, because the particles in solids vibrate around fixed points.

When a solid is heated until its particles vibrate so rapidly that they are no longer held in fixed positions, the solid melts.

Melting point is the temperature at which a solid changes to a liquid. The melting and freezing point of a substance are at the same temperature.

In general, ionic solids tend to have relatively high melting points, while molecular solids tend to have relatively low melting points.

Most solids are crystalline

The particles are arranged in a pattern known as a crystal lattice

The smallest subunit of a crystal lattice is the unit cell

Some solids lack an ordered internal structure and are called amorphous solids.

7 0

3 years ago

How many grams of carbon dioxide will be produced if 76.4 grams of

goldfiish [28.3K]

Answer:

Assuming that all of the oxygen is used up, 1.53×4111.53×411 or 0.556 moles of C2H3Br3 are required. Because there are only 0.286 moles of C2H3Br3 available, C2H3Br3 is the limiting reagent.

Limiting Reagent What is the limiting reagent if 76.4 grams of C2H3Br3 were reacted with 49.1 grams of O2? C2H3Br3 + 11O2 → 8CO2 + 6H2O + 6Br2 SOLUTION Using Approach 1: A. 76.4g × (1 mol/ 266.72 g) = 0.286 moles C2H3Br3 49.1g × (1 mole/ 32 g) = 1.53 moles O2 B.

Explanation:

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https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Map%3A_Introductory_Chemistry_(Tro)/08%3A_Quantities_in_Chemical_Reactions/8.04%3A_Limiting_Reactant_and_Theoretical_Yield

5 0

3 years ago

Nitrogen dioxide is used industrially to produce nitric acid, but it contributes to acid rain and photochemical smog. What volum

Strike441 [17]

Answer:

35.41 L

Explanation:

Given, Volume of Copper = 4.84 cm³

Density = 8.95 g/cm³

Considering the expression for density as:

$Density=\frac {Mass}{Volume}$

So,

So, Mass= Density Volume = 8.95 g/cm³ 4.84 cm³ = 43.318 g

Mass of copper = 43.318 g

Molar mass of copper = 63.546 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{43.318\ g}{63.546\ g/mol}$

Moles of copper = 0.6817 moles

Given, Volume of nitric acid solution = 227 mL = 227 cm³

Density = 1.42 g/cm³

Considering the expression for density as:

$Density=\frac {Mass}{Volume}$

So,

So, Mass= Density Volume = 1.42 g/cm³ 227 cm³ = 322.34 g

Also, Nitric acid is 68.0 % by mass. So,

Mass of nitric acid = $\frac {68}{100}\times 322.34\ g$ = 219.1912 g

Molar mass of nitric acid = 63.01 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{219.1912\ g}{63.01\ g/mol}$

Moles of nitric acid = 3.4786 moles

According to the reaction,

$Cu_{(s)}+4HNO_3_{(aq)}\rightarrow Cu(NO_3)_2_{(aq)} + 2NO_2_{(g)} + 2H_2O_{(l)}$

1 mole of copper react with 4 moles of nitric acid

Thus,

0.6817 moles of copper react with 4*0.6817 moles of nitric acid

Moles of nitric acid required = 2.7268 moles

Available moles of nitric acid = 3.4786 moles

Limiting reagent is the one which is present in small amount. Thus, nitric acid is present in large amount, copper is the limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of copper on reaction forms 2 moles of nitrogen dioxide

So,

0.6817 mole of copper on reaction forms 2*0.6817 moles of nitrogen dioxide

Moles of nitrogen dioxide = 1.3634 moles

Given:

Pressure = 724 torr

The conversion of P(torr) to P(atm) is shown below:

$P(torr)=\frac {1}{760}\times P(atm)$

So,

Pressure = 724 / 760 atm = 0.9526 atm

Temperature = 28.2 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (28.2 + 273.15) K = 301.35 K

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9526 atm × V = 1.3634 mol × 0.0821 L.atm/K.mol × 301.35 K

⇒V = 35.41 L

3 0

3 years ago