Answer;
= g/cm^3
Explanation;
-Density is the mass per volume of a substance, or the weight per volume of an object. The Metric System is a type of system of measuring that uses only one root word for each basic dimension, such as, -gram for mass or -meter for distance.
-Metric system densities are usually in the units of mass per volume, such as g/cm^3 (gram per cubic centimeter). Where grams represents the units of mass and cm^3 represents the volume.
Uno de los componentes del “sal de uvas”, es el bicarbonato de sodio, es un sólido cristalino blanco, debido a su capacidad de neutralizar el exceso de ácido clorhídrico del estómago.
La acidez estomacal es ocasionada por un exceso en la producción de ácido clorhídrico (HCl) en el jugo gástrico del estómago. Uno de los componentes del “sal de uvas”, utilizada como antiácido, es el bicarbonato de sodio (NaHCO₃), es un sólido cristalino blanco, que neutraliza el exceso de ácido clorhídrico. La reacción de neutralización es:
HCl(aq) + NaHCO₃(aq) ⇒ NaCl(aq) + H₂O(l) + CO₂(g)
Uno de los componentes del “sal de uvas”, es el bicarbonato de sodio, es un sólido cristalino blanco, debido a su capacidad de neutralizar el exceso de ácido clorhídrico del estómago.
Puedes aprender más sobre neutralización aquí: brainly.com/question/23261152
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The normality of the H₂SO₄ that reacted with 25cc of 5 % NaOH solution is 1.1 N.
<h3>What is the molarity of 5% NaOH?</h3>
The molarity of 5% NaOH is 1.32 M
25 cc of NaOH neutralized 30cc of H₂SO₄ solution.
Equation of reaction is given below:
- 2 NaOH + H₂SO₄ ---> Na₂SO₄ + 2 H₂O
Molarity of H₂SO₄ = 1.32 x 1 x 25/(30 x 2) = 0.55 M
- Normality = Molarity × moles of H⁺ ions per mole of acid
moles of H⁺ ions per mole of H₂SO₄ = 2
Normality of H₂SO₄ = 0.55 x 2 = 1.1 N
In conclusion, the normality of an acid is determined from the molarity and the moles of H⁺ ions per mole of acid.
Learn more about normality at: brainly.com/question/22817773
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Answer:
b)15.0°C
Explanation:
Specific Heat of Water=4.2 J/g°C
This means, that 1 g of Water will take 4.2 J of energy to increase its temperature by 1°C.
∴80 g Water will take 80×4.2 J of energy to increase its temperature by 1°C.
80×4.2 J=336 J
Total Energy Provided=1680 J
The temperature increase=\frac{\textrm{Total energy required}}{\textrm{energy required to increase temperature by one degree}}
Temperature increase=
=5°C
Initial Temperature =10°C
Final Temperature=Initial + Increase in Temperature
=10+5=15°C
