3. Rebecca puts her backpack on a fan cart and turns the fan on. She then measures the speed each second afterward. At 10 second
1 answer:
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Explanation:
- It is known that the amount of heat necessary to raise the temperature of 1 gram of a substance by
is known as specific heat.
Since, q =
So, larger is the specific heat of a substance less will be the change in its temperature.
Therefore, olive oil has less specific heat as compared to water. This means that olive oil would get hotter.
- Similarly, the specific heat of gold is lesser than the given materials or metals. Hence, gold will requires less heat to rise its temperature.
As a result, water present in gold will heat readily.
- As the relation between heat and specific heat is as follows.
q =
Therefore, calculate the amount of heat required by the water as follows.
q =
=
= 33440 J
or, = 33.44 kJ (as 1 kJ = 1000 J)
Thus, 33.44 kJ heat would it take to raise the temperature of 100.0 g of water from to
.
Answer:
Explanation:
a. NaCl+ AgNO3-----> NaNO3+ AgCl
b. Số mol của NaCl= 0,2*0,5= 0,1 (mol)
------> Số mol của kết tủa AgCl tạo thành= 0,1 mol (dựa vào phương trình hóa học)
-----> Khối lượng của kết tủa AgCl tạo thành= 0,1*143,5=14,35(g)
c. Số mol của AgNO3= số mol của NaCl= 0,1 (mol)
------> Nồng độ mol của dd AgNO3 đã tham gia phản ứng= =0,25(M)
Answer:
- Mass of monobasic sodium phosphate = 1.857 g
- Mass of dibasic sodium phosphate = 1.352 g
Explanation:
<u>The equilibrium that takes place is:</u>
H₂PO₄⁻ ↔ HPO₄⁻² + H⁺ pka= 7.21 (we know this from literature)
To solve this problem we use the Henderson–Hasselbalch (<em>H-H</em>) equation:
pH = pka +
In this case [A⁻] is [HPO₄⁻²], [HA] is [H₂PO₄⁻], pH=7.0, and pka = 7.21
If we use put data in the <em>H-H </em>equation, and solve for [HPO₄⁻²], we're left with:
From the problem, we know that [HPO₄⁻²] + [H₂PO₄⁻] = 0.1 M
We replace the value of [HPO₄⁻²] in this equation:
0.616 * [H₂PO₄⁻] + [H₂PO₄⁻] = 0.1 M
1.616 * [H₂PO₄⁻] = 0.1 M
[H₂PO₄⁻] = 0.0619 M
With the value of [H₂PO₄⁻] we can calculate [HPO₄⁻²]:
[HPO₄⁻²] + 0.0619 M = 0.1 M
[HPO₄⁻²] = 0.0381 M
With the concentrations, the volume and the molecular weights, we can calculate the masses:
- Molecular weight of monobasic sodium phosphate (NaH₂PO₄)= 120 g/mol.
- Molecular weight of dibasic sodium phosphate (Na₂HPO₄)= 142 g/mol.
- mass of NaH₂PO₄ = 0.0619 M * 0.250 L * 120 g/mol = 1.857 g
- mass of Na₂HPO₄ = 0.0381 M * 0.250 L * 142 g/mol = 1.352 g