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3 years ago

What is the frame of reference?

Chemistry

1 answer:

erma4kov [3.2K]3 years ago

5 0

Answer:

<em>Frame of Reference</em>

The fixed background surrounding an object, assumed to be at rest and in a fixed position.

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Calculate the volume of 12.0 g of helium at 100° C and 1.2 atm.

satela [25.4K]

Answer:

the answer I think is 76.5L

7 0

3 years ago

Please help 100 points.

Alexandra [31]

A. Snails and crayfish would not survive in Birch Lake at a pH of 6.2.

B. Once they die out, the walleye will disappear as well.

Walleye: pH 6.0 — pH 9.0

Crayfish: pH 6.5 — pH 8.0

Snails: pH 6.4 — pH 8.0

At pH = 7.3, all three can survive.

At pH = 6.2, snails and crayfish will die out.

Then, the walleye will disappear because <em>their food supply has vanished</em>.

6 0

3 years ago

Calculate the pH at the equivalence point when 22.0 mL of 0.200 M hydroxylamine, HONH2, is titrated with 0.15 M HCl. (Kb for HON

solniwko [45]

Answer:

pH = 3.513

Explanation:

Hello there!

In this case, since this titration is carried out via the following neutralization reaction:

$HONH_2+HCl\rightarrow HONH_3^+Cl^-$

We can see the 1:1 mole ratio of the acid to the base and also to the resulting acidic salt as it comes from the strong HCl and the weak hydroxylamine. Thus, we first compute the required volume of HCl as shown below:

$V_{HCl}=\frac{22.0mL*0.200M}{0.15M}=29.3mL$

Now, we can see that the moles of acid, base and acidic salt are all:

$0.0220L*0.200mol/L=0.0044mol$

And therefore the concentration of the salt at the equivalence point is:

$[HONH_3^+Cl^-]=\frac{0.0044mol}{0.022L+0.0293L} =0.0858M$

Next, for the calculation of the pH, we need to write the ionization of the weak part of the salt as it is able to form some hydroxylamine as it is the weak base:

$HONH_3^++H_2O\rightleftharpoons H_3O^++HONH_2$

Whereas the equilibrium expression is:

$Ka=\frac{[H_3O^+][HONH_2]}{[HONH_3^+]}$

Whereas Ka is computed by considering Kw and Kb of hydroxylamine:

$Ka=\frac{Kw}{Kb}=\frac{1x10^{-14}}{9.10x10^{-9}} \\\\Ka=1.10x10^{-6}$

So we can write:

$1.10x10^{-6}=\frac{x^2}{0.0858-x}$

And neglect the x on bottom to obtain:

$1.10x10^{-6}=\frac{x^2}{0.0858}\\\\x=\sqrt{1.10x10^{-6}*0.0858}=3.07x10^{-4}M$

And since x=[H3O+] we obtain the following pH:

$pH=-log(3.07x10^{-4})\\\\pH=3.513$

Regards!

4 0

3 years ago

Given that a 11.00 g milk bar chocolate bar contains 7.500 g of sugar, calculate the percentage of sugar present in 11.00 g of m

alukav5142 [94]

Answer:

68.18%

Explanation:

The question asking for the percentage of sugar present in milk bar chocolate. To find the percentage of a certain molecule, you have to divide the mass of the molecule with the total mass of the product. The calculation will be:

sugar percentage = sugar weight / milk bar chocolate weight * 100%

sugar percentage = 7.5g / 11g * 100%= 68.18%

7 0

3 years ago

A bond formed when atoms transfer elentron is a bond

lubasha [3.4K]

Ionic bonds involve a cation and an anion. The bond is formed when an atom, typically a metal, loses an electron or electrons, and becomes a positive ion, or cation. Another atom, typically a non-metal, is able to acquire the electron(s) to become a negative ion, or anion.

One example of an ionic bond is the formation of sodium fluoride, NaF, from a sodium atom and a fluorine atom. In this reaction, the sodium atom loses its single valence electron to the fluorine atom, which has just enough space to accept it. The ions produced are oppositely charged and are attracted to one another due to electrostatic forces.

5 0

4 years ago