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3 years ago

What can humans do to reduce climate change

Chemistry

2 answers:

yanalaym [24]3 years ago

6 0

Answer:

Ten possibilities for staving off catastrophic climate change ... What can one person, or even one nation, do on their own to slow and reverse climate change? ... could reduce greenhouse gas emissions to safer levels—there are ... mode of transport that does not require anything other than human energy.

Explanation:

abruzzese [7]3 years ago

3 0

They can reduce greenhouse gas

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if the density of copper is 8.9g/cm^3 and the Volume of the sample is 23.4 mL what would be the mass?

Black_prince [1.1K]

Answer:

The mass of copper = 208.26 grams.

Explanation:

Density = mass / volume

8.9 = mass / 23.4

mass = 8.9 * 23.4

= 208.26 grams.

8 0

3 years ago

101.325 kPa = 760 mmHg. So, 25.0 kPa is

gogolik [260]

101.325 kPa = 760 mmHg.
So, 25.0 kPa is 187.515 mmHg

5 0

3 years ago

The number of covalent bonds that an atom can make is determined by

diamong [38]

The number of covalent bonds that an atom can make is determined by the no. of electrons needed to form a duplet or octet of electron by each of the atom.

5 0

4 years ago

At a certain temperature, 0.760 mol SO3 is placed in a 1.50 L container. 2SO3(g) = 2SO2(g) + O2(g)At equilibrium, 0.130 mol O2 i

olganol [36]

Answer:

$K_c=0.0867$

Explanation:

Moles of SO₃ = 0.760 mol

Volume = 1.50 L

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.760}{1.50\ L}$

[SO₃] = 0.5067 M

Considering the ICE table for the equilibrium as:

$\begin{matrix} & 2SO_3_{(g)} & \rightleftharpoons & 2SO_2_{(g)} & + & O_2_{(g)}\\At\ time, t = 0 & 0.5067 &&0&&0 \\ At\ time, t=t_{eq} & -2x &&2x&&x \\----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:- &0.5067-2x&&2x&&x\end{matrix}$

Given:

Equilibrium concentration of O₂ = 0.130 mol

Volume = 1.50 L

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.130}{1.50\ L}$

[O₂] = x = 0.0867 M

[SO₂] = 2x = 0.1733 M

[SO₃] = 0.5067-2x = 0.3334 M

The expression for the equilibrium constant is:

$K_c=\frac {[SO_2]^2[O_2]}{[SO_3]^2}$

$K_c=\frac{(0.3334)^2\times 0.0867}{(0.3334)^2}$

$K_c=0.0867$

5 0

3 years ago

1. 3.311 * 10^24 particles=_______________moles? 2. 126 grams of H2O =______________________moles of moles? whats the answers

Vladimir [108]

1. You can use Avogadro’s number, 6.022x10^23 atoms/mole, to answer this one.

(3.311x10^24/6.022x10^23) = 5.498 moles of substance

2. H2O has a formula weight approximately equal to 18 grams. Dividing the given amount by the formula weight of water will tell us the number of moles present.

126/18 = 7 moles H2O

4 0

3 years ago