Answer:
0.4762 J/g°C.
Explanation:
<em>The amount of heat released to water = Q = m.c.ΔT.</em>
where, m is the mass of water (m = 15.0 g).
c is the specific heat capacity of water = ??? J/g°C.
ΔT is the temperature difference = (final T - initial T = 37.0°C - 30.0°C = 7.0°C).
<em>∴ The specific heat capacity of water = c = Q/m.ΔT</em> = (50.0 J)/(15.0 g)(7.0°C) = <em>0.4762 J/g°C.</em>
⇒Answer:
When the pH sensor hits pH=7.
⇒Explanation:
Because pH=7 is the indicator that the acid and alkali have been neutralized.
Answer:
pulmonary vein
Explanation:
The pulmonary vein is the only vein in the body that carries oxygen rich blood.
The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL
<h3>Balanced equation </h3>
2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O
From the balanced equation above,
- The mole ratio of the acid, H₃PO₄ (nA) = 2
- The mole ratio of the base, Ca(OH)₂ (nB) = 3
<h3>How to determine the volume of Ca(OH)₂ </h3>
- Molarity of acid, H₃PO₄ (Ma) = 0.390 M
- Volume of acid, H₃PO₄ (Va) = 24.5 mL
- Molarity of base, Ca(OH)₂ (Mb) = 0.279 M
- Volume of base, Ca(OH)₂ (Vb) =?
MaVa / MbVb = nA / nB
(0.39 × 24.5) / (0.279 × Vb) = 2/3
9.555 / (0.279 × Vb) = 2/3
Cross multiply
2 × 0.279 × Vb = 9.555 × 3
0.558 × Vb = 28.665
Divide both side by 0.558
Vb = 28.665 / 0.558
Vb = 51.4 mL
Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL
Learn more about titration:
brainly.com/question/14356286
Answer:
The limiting reactant is the 6.279 g of
Explanation:
We have to start with the <u>reaction</u> between sodium carbonate () and the Nickel (II) Chloride (), so:
We will have a <u>double replacement reaction</u>. Now we have to <u>balance</u> the reaction, so:
The next step is the <u>calculation of the moles for each reactive</u>. For we have use the <u>molarity equation</u>:
For the calculation of moles of we have to use the <u>molar mass</u> of the compound (129.59 g/mol):
The next step is the division of each mole value by the <u>coefficient of each reactive</u> of the balance reaction. In this case <u>we have "1" for each reactive</u>, so:
The final step is to <u>choose the smallest value</u>. In this case is the value that correspond to . Therefore is the limiting reactive.