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mariarad [96]
2 years ago
13

Which substance is not a solid at 20˚C and one atmosphere of pressure?

Chemistry
1 answer:
levacccp [35]2 years ago
3 0
Choose A, Kr is noble gas
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PLEASE HELP ASAP!!!<br> K+ has how many electrons? <br> -19<br> -18<br> -20<br> -none of these
Y_Kistochka [10]

Answer:

Its 19.

Explanation:

K (potassium) has 19 electron according to the atomic number in the periodic table. The atomic number tells you the number of protons and electrons.

8 0
3 years ago
2
melamori03 [73]

Answer:

D. liquid

Explanation:

Matter, which is any substance that has weight and occupies space, can exists in three states namely: solid, liquid and gaseous. In this question, water is being described as matter.

According to this question, a sample of water of volume 2litres is shaped like a cylinder in one container and shaped like a rectangle in another larger container. Based on this observation, the water sample is in a LIQUID STATE because a liquid has no indefinite shape but takes the shape of its container.

Hence, this water sample, takes the shape of its cylindrical container in the first scenario and shape of its rectangular container in the second scenario.

8 0
3 years ago
Why did the French set up trading posts in North America?
erma4kov [3.2K]

Answer: Their main aim was to provide goods, specifically furs and sugar exportation.

Explanation:

3 0
3 years ago
What are two ways in which density has practical importance?
Marta_Voda [28]
<span>density is how much matter is packed into one area. so the higher density the more likely it is to be heavier. for example, cork has a very low density and iron has a higher density.</span>
6 0
3 years ago
1. 51.2 g of NaClO (pKa of HClO is 7.50) and 25.3 g of KF (pKa of of HF is 3.17) were combined in enough water to make 0.250 L o
Andrews [41]

Answer:

* pH=11.0

* [HClO]_{eq}=9.32x10^{-4}M

* [HF]_{eq}=5.07x10^{-6}M

Explanation:

Hello,

At first, the molarities of NaClO and KF are computed as shown below:

M_{NaClO}=\frac{51.2g*\frac{1mol}{74.45g} }{0.250L}=2.75M \\\\M_{KF}=\frac{25.3g*\frac{1mol}{58.1g} }{0.250L} =1.74M

Now, since both NaClO and KF are ionic, one proposes the dissociation reactions as:

NaClO\rightarrow Na^++ClO^-\\KF\rightarrow K^++F^-

In such a way, by writing the formation of HClO and HF respectively, due to the action of water, we obtain:

ClO^-+H_2O\rightleftharpoons HClO+OH^-\\F^-+H_2O\rightleftharpoons HF+OH^-\\

Now, the presence of OH⁻ immediately implies that Kb for both NaClO and KF must be known as follows:

Kb=\frac{Kw}{Ka} \\Kb_{NaClO}=\frac{1x10^{-14}}{10^{-7.50}}=3.16x10^{-7}\\Kb_{KF}=\frac{1x10^{-14}}{10^{-3.17}}=1.48x10^{-11}

In such a way, the law of mass action for each case is:

Kb_{NaClO}=3.16x10^{-7}=\frac{[HClO][OH]^-}{[ClO^-]}=\frac{(x_{ClO})(x_{ClO})}{2.75-x_{ClO}}\\Kb_{KF}=1.48x10^{-11}=\frac{[HF][OH]^-}{[F^-]}=\frac{(x_{F})(x_{F})}{1.74-x_{F}}

Now, by solving for the change x for both the ClO⁻ and F⁻ equilibriums, we obtain:

x_{ClO}=9.32x10^{-4}M\\x_{F}=5.07x10^{-6}M

Such results equal the concentrations of OH⁻ in each equilibrium, thus, the total concentration of OH⁻ result:

[OH^-]_{tot}=9.32x10^{-4}M+5.07x10^{-6}M=9.37x10^{-4}M

With which the pOH is:

pOH=-log(9.37x10^{-4})=3.03

And the pH:

pH=14-pOH=11.0

In addition, the equilibrium concentrations of HClO and HF equals the change x for each equilibrium as:

[HClO]_{eq}=x_{ClO}=9.32x10^{-4}M

[HF]_{eq}=x_{F}=5.07x10^{-6}M

Best regards.

6 0
3 years ago
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