Which of the following is a way that trees have been negatively impacted by human use?

A student wonders whether a piece of jewelry is made of pure silver. She determines that its mass is 3.25 g. Then she drops it i

nydimaria [60]

Answer:

The jewelry is 2896.54_Kg/m^3 less dense than pure silver

Explanation:

Density of jewellery = (mass of jewellery) ÷ (volume of jewellery)

=3.25g ÷ 0.428mL = 0.00325Kg÷0.000000428m^3 = 7583.46Kg/m^3

The density of silver is 10490_Kg/m^3 which is (10490 - 7583.46) 2896.54_Kg/m^3 more dense than the jewellery

The density of Silver [Ag]

The weight of Silver per cubic centimeter is 10.49 grams or the weight of silver per cubic meter is 10490 kilograms, that is the density of silver is 10490 kg/m³; at 20°C (68°F or 293.15K) at a pressure of one atmospheres.

7 0

3 years ago

A concentrated aqueous solution of Pb(NO3)2 is slowly added to 1.0 L of a mixed aqueous solution containing 0.010 M Na2CrO4 and

Oduvanchick [21]

Answer:

This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M

Explanation:

Step 1: Data given

Molarity of Na2CrO4 = 0.010 M

Molarity of NaBr = 2.5 M

Ksp(PbCrO4) = 1.8 * 10^–14

Ksp(PbBr2) = 6.3 * 10^–6

Step 2: The balanced equation

PbCrO4 →Pb^2+ + CrO4^2-

PbBr2 → Pb^2+ + 2Br-

Step 3: Define Ksp

Ksp PbCrO4 = [Pb^2+]*[CrO4^2-]

1.8*10^-14 = [Pb^2+] * 0.010 M

[Pb^2+] = 1.8*10^-14 /0.010

[Pb^2+] = 1.8*10^-12 M

The minimum [Pb^2+] needed to precipitate PbCrO4 is 1.8*10^-12 M

Ksp PbBr2 = [Pb^2+][Br-]²

6.3 * 10^–6 = [Pb^2+] (2.5)²

[Pb^2+] = 1*10^-6 M

The minimum [Pb^2+] needed to precipitate PbBr2 is 1*10^-6 M

This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M

5 0

3 years ago

What effect does cutting down forests have on the carbon dioxide levels in the atmosphere?

Mnenie [13.5K]

Https://worldcuprussiai.de/ http://www.frankieballard.com/forum/soccer-brazil-vs-mexico-live-stream-online-free-2nd-round-117876

5 0

3 years ago

after a chemistry student made a AgNO3(small 3 at bottom) solution., she wanted to determine the molar concentration of it. if 2

Alexxandr [17]

Molar concentration = (numbet of mol Solute)/ ( volume Solution)

1) Finding
the number of the mol solute

$22.0 g AgNO3 * \frac{1 mol AgNO3}{169.91 g AgNO3} = 0.129 mol AgNO3

2) 725 ml = 0.725 L


$

$3) Molarity = \frac{number/ mol/solute}{Volume/solution} = \frac{0.129}{0.725} 

$

$Molarity= \frac{0.178 mol}{L} , 

or 0.178 M
$

3 0

4 years ago

3. How much energy is needed to raise 45 grams of water from 40°C to 115 °C?

Dafna1 [17]

Answer:

Q = 114349.5 J

Explanation:

Hello there!

In this case, since this a problem in which we need to calculate the total heat of the described process, it turns out convenient to calculate it in three steps; the first one, associated to the heating of the liquid water from 40 °C to 100 °C, next the vaporization of liquid water to steam at constant 100 °C and finally the heating of steam from 100 °C to 115 °C. In such a way, we calculate each heat as shown below:

$Q_1=45g*4.18\frac{J}{g\°C}*(100\°C-40\°C)=11286J\\\\Q_2=45g* 2260 \frac{J}{g} =101700J\\\\Q_3=45*2.02\frac{J}{g\°C}*(115\°C-100\°C)=1363.5J$

Thus, the total energy turns out to be:

$Q_T=11286J+101700J+1363.5J\\\\Q_T=114349.5J$

Best regards!

5 0

3 years ago