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3 years ago

1. Give the best advice on being understanding but kinda flirty to a friend that is a girl.

1 answer:

5 0

Being understanding is flirty/attractive on its own. listening to her and asking questions bout the situation really shows her that you care and will make you attractive, also maybe inviting to go out to get her mind off things or getting her her fav candy & some other cute little things really will show her that you like her. just be nice to her and listen to what she has to say.

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reaction 1:CCl4⟶CH4 reaction 2:N2+3H2⟶2H3N reaction 3:2H2+2O2⟶2H2O+Au Which of the chemical reactions are NOT possible according

olga nikolaevna [1]

Clutch Prep

Ch.2 - Atoms & ElementsSee all chapters

Atomic Theory

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LearnAdditional Practice

Solution: Which of the following chemical reactions is/are NOT possible according to Dalton's atomic theory?a. reaction 1: CCl4 → CH4b. reaction 2: N2 + 3H2 → 2NH3c. reaction 3: 2H2 + O2 → 2H2O + Au

Problem

Which of the following chemical reactions is/are NOT possible according to Dalton's atomic theory?

a. reaction 1: CCl4 → CH4

b. reaction 2: N2 + 3H2 → 2NH3

c. reaction 3: 2H2 + O2 → 2H2O + Au

3 0

3 years ago

which tool was mostly likely used in a procedure if the lab report shows that approximately 300 ml of water was used

DiKsa [7]

it is most likely a beaker, beacuse 300ml is quite a large volume. Otherwise, it would be a measuring cylinder or pippette

7 0

3 years ago

A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.

irga5000 [103]

Answer:

1. pH = 1.23.

2. $H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

$H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+$

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the pKa is:

$pKa=-log(Ka)=-log(5.90x10^{-2})=1.23$

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

$pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23$

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

$n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol$

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

$H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Which is also shown in net ionic notation.

Best regards!

4 0

3 years ago

Sometimes people do not want to change their belief regardless of the evidence.

bagirrra123 [75]

Answer:

True

Explanation:

Ex. People against vaccines tend to believe crazy theories over facts because they think they'll end up brainwashing them.

6 0

3 years ago

How many protons and neutrons does Polonium-211 have?

Stells [14]

Answer:

Protons: 84

Neutrons: 127

Explanation:

The number of protons is the atomic number. The number of neutrons can be found by subtracting the proton number (84) from the isotope number (211)

3 0

3 years ago