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2 years ago

1. At what temperature dose water freeze

Chemistry

1 answer:

Rus_ich [418]2 years ago

8 0

Answer:

1. water will freeze at a temperature below 32 degrees fahrenheit 0 degree celsius.

2. Ice will melt at a temperature above 32 degrees fahrenheit 0 degrees celsius.

3. water boils at 212 degrees fahrenheit or 100 degrees celsius.

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Which agents cause both chemical and physical weathering

shutvik [7]

Weathering is the process by which rocks are broken down by either physical or chemical means. Physical weathering involves physical factors that cause mechanical break down of rocks while chemical weathering involves chemical reactions between the contents of the rocks and other factors such air (oxygen). Agents of weathering are the factors that cause or enhance both physical and chemical weathering. Some of these factors cause both physical and chemical weathering these includes water and temperature.

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3 years ago

The symbols from the list and match them with the names of the

stich3 [128]

Be - Beryllium
S - sulfur
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C - Carbon
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3 years ago

Please help me with this

Sophie [7]

Answer:

5.0 38 84.0 749.7 528.0 729.0 738.9 739.0

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2 years ago

What else can copper react with?

FinnZ [79.3K]

Answer:

copper

Explanation:

7 0

3 years ago

Read 2 more answers

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