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nadya68 [22]
3 years ago
8

The client's lab values are sodium 166 mEq/L, potassium 5.0 mEq/L, chloride 115 mEq/L, and bicarbonate 35 mEq/L. What condition

is this client likely to have, judging by anion gap?
Chemistry
1 answer:
Fantom [35]3 years ago
5 0

Answer:

The anion value of 21mEq/L suggest the client will have metabolic acidosis

Explanation:

Anion gap is the interval or difference between measured cations (Sodium and Potassium) with the sum of measured anions (chloride and bicarbonate)

Using the above mentioned data

(166 + 5) - (115 + 35)  =  21mEq/L

Therefore the Anion gap = 21mEq/L

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Write the balanced standard combustion reaction for the c5h7on3s2
AlexFokin [52]

Answer:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Explanation:

When C5H7ON3S2 under go Combustion, the following are obtained as illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

Now, let us balance the equation. This is illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

There are 3 atoms of N on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of C5H7ON3S2 and 6 in front of N2 as shown below:

4C5H7ON3S2 + O2 —> CO2 + H2O + 6N2 + S2

There are 20 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 20 in front of CO2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + H2O + 6N2 + S2

There are 28 atoms on H on the left side and 2 atoms on the right side. It can be balance by putting 14 in front of H2O as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + S2

There are 8 atoms of S on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of S2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, there are a total of 54 atoms of O on the right side and 6 atoms on the left side

It can be balance by putting 25 in front of O2 as shown below:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, we can see that the equation is balanced.

4 0
2 years ago
In an acid-base titration, a student uses 21.35 mL of 0.150 M NaOH to neutralize 25.00 mL of H2SO4. How many moles of acid are i
GalinKa [24]

Answer: There are 0.006 moles of acid in the flask.

Explanation:

Given: V_{1} = 21.35 mL,        M_{1} = 0.150 M

V_{2} = 25.0 mL,           M_{2} = ?

Formula used to calculate molarity of H_{2}SO_{4} is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\0.15 M \times 21.35 mL = M_{2} \times 25.0 mL\\M_{2} = 0.1281 M

As molarity is the number of moles of a substance present in a liter of solution.

Total volume of solution = V_{1} + V_{2}

= 21.35 mL + 25.0 mL

= 46.36 mL  (1 mL = 0.001 L)

= 0.04636 L

Therefore, moles of acid required are calculated as follows.

Molarity = \frac{no. of moles}{Volume (in L)}\\0.1281 M = \frac{no. of moles}{0.04635 L}\\no. of moles = 0.006 mol

Thus, we can conclude that there are 0.006 moles of acid in the flask.

3 0
3 years ago
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