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olchik [2.2K]
3 years ago
8

The name of an oxyacid has the suffix -ous acid. What is the suffix of the oxyanion?

Chemistry
1 answer:
Dima020 [189]3 years ago
6 0

Answer:

-ate

Explanation:

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A solution that is 0.20 m in hcho2 and 0.15 m in nacho2 find ph
Mashutka [201]
We are given
0.2 M HCHO2 which is formic acid, a weak acid
and
0.15 M NaCHO2 which is a salt which can be formed by reacting HCHO2 and NaOH

The mixture of the two results to a basic buffer solution
To get the pH of a base buffer, we use the formula
pH = 14 - pOH = 14 - (pKa - log [salt]/[base])

We need the pKa of HCO2
From, literature, pKa = 1.77 x 10^-4
Substituting into the equation
pH = 14 - (1.77 x 10^-4 - log 0.15/0.2)
pH = 13.87

So, the pH of the buffer solution is 13.87
A pH of greater than 7 indicates that the solution is basic and a pH close to 14 indicates high alkalinity. This is due to the buffering effect of the salt on the base.
5 0
3 years ago
PLEASEEE HELPPP MEEE
andrezito [222]

Answer:

yes the one that is circled is correct

Explanation:

4 0
3 years ago
Read 2 more answers
Given that kb for c6h5nh2 is 1.7 × 10-9 at 25 °c, what is the value of ka for c6h5nh3 at 25 °c?
lina2011 [118]

Answer: k_a for C_6H_5NH_3^+ at 25°C is 0.588\times 10^{-4}

Explanation: We are given k_b of C_6H_5NH_2 at 25°C which is 1.7\times 10^{-9}

To calculate the k_a of C_6H_5NH_3^+, we use the formula:

k_w=k_a\times k_b

k_w\text{ at }25^o=1\times 10^{-14}

Putting values in above equation, we get:

1\times 10^{-14}=k_a\times (1.7\times 10^{-9})\\k_a=0.588\times 10^{-5}

7 0
3 years ago
2CO + O2  2CO2
murzikaleks [220]

Answer:

75 L Co2

37.5 L o2

Explanation:

welcome.

3 0
3 years ago
Promethium-147 is sometimes used in luminescent paint. It has a half-life of 956.3 days. If 250 grams (g) of promethium-147 is u
Fantom [35]

Answer:

% = 76.75%

Explanation:

To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:

A = A₀ e^(-kt)   (1)

Where:

A and A₀: concentrations or mass of the compounds, (final and initial)

k: constant decay of the compound

t: given time

Now to get the value of k, we should use the following expression:

k = ln2 / t₁/₂   (2)

You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.

Now, let's calculate k:

k = ln2 / 956.3

k = 7.25x10⁻⁴ d⁻¹

With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:

A = 250 e^(-7.25x10⁻⁴ * 365)

A = 250 e^(-0.7675)

A = 191.87 g

However, the question is the percentage left after 1 year so:

% = (191.87 / 250) * 100

<h2>% = 76.75%</h2><h2>And this is the % of isotope after 1 year</h2>
3 0
3 years ago
Read 2 more answers
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