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Mekhanik [1.2K]
3 years ago
14

Substances A and B are combined. When combined, substance A, which is a black solid, sinks to the bottom of substance B which is

a clear liquid. Based on this situation, which of the following is true.
A. Substances A and B formed a mixture
B. It is impossible to tell without more information
C. Substances A and B formed a compound
D. Substances A and B formed a new element
Chemistry
2 answers:
Makovka662 [10]3 years ago
4 0

Answer:

c thats just what i do

Explanation:

Kamila [148]3 years ago
4 0
I think it would be A
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How many grams of oxygen are produced when 1.05 moles of hydrogen gas is<br> produced?
vivado [14]

Answer:

1.058337 grams of hydrogen and 2H2 + O2 ==> 2H2O hydrogen peroxide

mols H2 = grams/molar mass

Using the coefficients in the balanced equation, convert mols H2 to mols O2.

Now convert mols O2 to grams. That's grams mols O2 x molar mass O2.

and it could also produce H2O water but no air but it could make other things

8 0
2 years ago
Classify the following aqueous solutions as: strong acid, weak acid, neutral, weak base, or strong base.
kirill [66]

Vinegar pH 3.2: Weak acid

Battery acid pH 0.5: Strong acid

Shampoo pH 7.0: Neutral

Ammonia pH 11.1 Strong base

4 0
1 year ago
What is the molar mass of C4H10
Phantasy [73]
Your answer would be 58.12g/mol ;)
7 0
4 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
3 &amp; 4 Help pleaseeeee
kvasek [131]
3 is C i believe, and as for 4 minerals are solid, natural.. that's about all I can say...
4 0
3 years ago
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