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3 years ago

Which of the following is not a greenhouse gas

Chemistry

1 answer:

Vinil7 [7]3 years ago

4 0

A- water. It’s the only one that isn’t a gas

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Which compound contains a radical

slamgirl [31]

The compound that contains a radical is called NaOH.

Explanation:

Radical, also called Free Radical, in chemistry, a fragment that contains at trivial one unpaired electron. Most molecules include even numbers of electrons, and the covalent chemical bonds including the atoms commonly within a molecule normally consist of pairs of electrons jointly shared by the atoms combined by the bond.

6 0

3 years ago

Read 2 more answers

The density of H2O2 is 1.407 g/mL, and the density of O2 is 1.428 g/L. How many liters of O2 can be made from 55 mL H2O2

balandron [24]

Explanation:

mass H2O2 = 55 mL(1.407 g/mL) = 80.85 g

molar mass H2O2 = 2(1.01 g/mol) + 2(16.00 g/mol) = 34.02 g/mol

moles H2O2 = 80.85 g/34.02 g/mol = 2.377 moles H2O2

For each mole of H2O2 you obtain 0.5 mole of O2 (see the equation).

moles O2 = 2.377 moles H2O2 (1 mole O2)/(2 moles H2O2) = 1.188 moles O2

Now, you need the temperature. If you are at STP (273 K, and 1.00 atm) then 1 mole of an ideal gas at STP has a volume of 22.4 L. Without temperature you are not really able to continue. I will assume you are at STP.

Volume O2 = 1.188 moles O2(22.4 L/mole) = 0.0530 L of O2.

which is 53 mL.

8 0

2 years ago

Which sequence of events is required to form a limestone cave where you can walk around and observe cave formations, such as sta

N76 [4]

Answer:

Which sequence of events is required to form a limestone cave where you can walk around and observe cave formations, such as stalactites? (Note: stalactites hang from the ceiling - they have to hold on tight to the roof.)

A geological sequence of events as involving the lowering of the water table to expose cave structures where stalactites and stalagmites form which is described as follows,

Explanation:

1. Acidic percolated water formed cavities of solution beneath the natural water table known as phreatic zone

2. After the passage of time there is a drop in the water table dropped forming caves from cavities

3. These caves, which are air filled voids that contains adequate environment for forming stalactites and stalagmites and where they are found

5 0

3 years ago

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

3 years ago

chemical properties a. include changes of state of a substance b. include mass and color c. include changes that alter the ident

8_murik_8 [283]

C. Include changes that alter the identity of a substance.

6 0

3 years ago