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3 years ago

What is the mass in grams of 3.05x10^25 gold atoms

1 answer:

4 0

9.96*10⁻²² is the answer, First you find the molar mass of gold, then you turn atoms to moles then moles to grams. There you go you're done!<span />

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A chemical reaction in which two or more substances undergo a chemical union to form a more complex substance is a _______ react

Georgia [21]

I believe the correct answer is B. Combination.

7 0

3 years ago

Read 2 more answers

The fact that same molecules of matter have been around for years is supported by

Elza [17]

Answer:

The law of definite proportions

Explanation:

The law of definite proportions states that atoms combine in a molecule in a specific molar ratio or specific stoichiometry. For example, it's proved that regardless of the quantity we take, two hydrogen atoms always combine with one oxygen atom to form a water molecule.

Similarly, ionic substances follow the same pattern. Since the net charge of ionic salts should be equal to 0 and each element has a definite number of valence electrons in its shell all the time, the ions combine in a way, so that cations balance the charge of anions.

Essentially, the law of definite proportions is applicable and will be applicable in the future, since we know that each element has a fixed number of valence electrons in its ground state.

6 0

3 years ago

A buffer contains 0.19 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. You may want to

Snezhnost [94]

Explanation:

It is known that $pK_{a}$ of propionic acid = 4.87

And, initial concentration of propionic acid = $\frac{0.19}{1.20}$

= 0.158 M

Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]

= 0.216 M

Now, in the given situation only propionic acid and sodium propionate are present .

Hence, pH = $pK_{a} + log(\frac{[salt]}{[acid]})$

= $4.87 + log \frac{0.216}{0.158}$

= 4.87 + log (1.36)

= 5.00

Therefore, when 0.02 mol NaOH is added then,

Moles of propionic acid = 0.19 - 0.02

= 0.17 mol

Hence, concentration of propionic acid = $\frac{0.17}{1.20 L}$

= 0.14 M

and, moles of sodium propionic acid = (0.26 + 0.02) mol

= 0.28 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

$\frac{0.28 mol}{1.20 L}$

= 0.23 M

Therefore, calculate the pH upon addition of 0.02 mol of NaOH as follows.

pH = $pK_{a} + log(\frac{[salt]}{[acid]})$

= $4.87 + log \frac{0.23}{0.14}$

= 4.87 + log (1.64)

= 5.08

Hence, the pH of the buffer after the addition of 0.02 mol of NaOH is 5.08.

Therefore, when 0.02 mol HI is added then,

Moles of propionic acid = 0.19 + 0.02

= 0.21 mol

Hence, concentration of propionic acid = $\frac{0.21}{1.20 L}$

= 0.175 M

and, moles of sodium propionic acid = (0.26 - 0.02) mol

= 0.24 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

$\frac{0.24 mol}{1.20 L}$

= 0.2 M

Therefore, calculate pH upon addition of 0.02 mol of HI as follows.

pH = $pK_{a} + log(\frac{[salt]}{[acid]})$

= $4.87 + log \frac{0.2}{0.175}$

= 4.87 + log (0.114)

= 4.98

Hence, the pH of the buffer after the addition of 0.02 mol of HI is 4.98.

7 0

3 years ago

In your own words, summarize the steps you took to complete the lab. Explain what the test (independent) variable was, and what

alexira [117]

We can’t answer this question without any information. There is no explanation of the lab, nor is there any data.

5 0

3 years ago

Maya made this picture to represent a chemical reaction:

natka813 [3]

Answer:

2) It is neither a synthesis reaction nor a decomposition reaction because two reactants form two products.

Explanation:

This is not a synthesis reaction, which takes place when two or more reactants combine to form one product. This is a single replacement reaction, where the cation of an ionic compound "trades places" with another element that becomes the cation in a new ionic compound.

8 0

3 years ago