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torisob [31]
2 years ago
5

What are the symptoms of high alkaline phosphatase?

Chemistry
1 answer:
Lesechka [4]2 years ago
5 0

Answer:

Itching.

Nausea and vomiting.

Weight loss.

Fatigue.

Weakness.

Jaundice.

Swelling and pain in your stomach.

Dark-colored urine and/or light-colored stool.

Explanation:

hope this helps

plz hit thanks

and plz mark brainliest

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You have a solution that is 0.02M formate (HCOO-) and 0.03M formic acid (HCOOH), which has a Ka of 1.8x10-4. What is the pH of t
lisov135 [29]

Answer:

3.6

Explanation:

Step 1: Given data

  • Concentration of formic acid: 0.03 M
  • Concentration of formate ion: 0.02 M
  • Acid dissociation constant (Ka): 1.8 × 10⁻⁴

Step 2: Calculate the pH

We have a buffer system formed by a weak acid (HCOOH) and its conjugate base (HCOO⁻). We can calculate the pH using the <em>Henderson-Hasselbach equation</em>.

pH = pKa +log\frac{[base]}{[acid]} = -log 1.8 \times 10^{-4} + log \frac{0.02}{0.03} = 3.6

6 0
3 years ago
An experiment produced 0.10 g CO2 with a volume of 0.056 L at STP. If the accepted density of CO2 at STP is 1.96 g/L, what is th
Mandarinka [93]
The experimental density of CO2 at STP is 0.10/0.056=1.78 g/L. The percent error equals to (1.96-1.78)/1.96*100%=9.18%. So the answer is 9.18%.
3 0
3 years ago
The density of solid cu is 8.96 g/cm3. how many atoms are present per cubic centimeter of cu?
svp [43]
Since there is 9.47 x 1021 atoms of copper in 1 gram, and 8.96 grams/cc is the density of copper, then 9.47 x 8.96 =84.85 x 1021 atoms of copper in the one cubic centimeter of copper.
3 0
3 years ago
How fast must a 56.5-g tennis ball travel in order to have a de Broglie wavelength that is equal to that of a photon of green li
nikklg [1K]

Answer:vv

v=2.17\times 10^{-26}\ m/s

Explanation:

The expression for the deBroglie wavelength is:

\lambda=\frac {h}{m\times v}

Where,  

\lambda is the deBroglie wavelength  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

m is the mass of = 56.5\ g=0.0565\ kg

v is the speed.

Wavelength = 5400 Å = 5400\times 10^{-10}\ m

Applying in the equation as:-

5400\times 10^{-10}=\frac{6.626\times 10^{-34}}{0.0565\times v}

v=\frac{331300000}{10^{34}\times \:1.5255}\ m/s

v=2.17\times 10^{-26}\ m/s

7 0
3 years ago
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Lady bird [3.3K]

Answer:0.004993 mols

Explanation:

4 0
3 years ago
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