Answer:
Both:
-They are both made up of cells embedded in an extracellular matrix. It is the nature of the matrix that defines the properties of these connective tissues.
Cartilage:
-Cartilage is thin, avascular, flexible and resistant to compressive forces.
-Cartilages are soft and flexible components present in ear, nose and joints.
Bone marrow:
-Bone is highly vascularised, and its calcified matrix makes it very strong.
-Bones are hard and tough which gives the structural framework of the skeleton in the body.
The specific heat of the metal, given the data from the question is 0.60 J/gºC
<h3>Data obtained from the question </h3>
The following data were obtained from the question:
- Mass of metal (M) = 74 g
- Temperature of metal (T) = 94 °C
- Mass of water (Mᵥᵥ) = 120 g
- Temperature of water (Tᵥᵥ) = 26.5 °C
- Equilibrium temperature (Tₑ) = 32 °C
- Specific heat capacity of the water (Cᵥᵥ) = 4.184 J/gºC
- Specific heat capacity of metal (C) =?
<h3>How to determine the specific heat capacity of the metal</h3>
The specific heat capacity of the sample of gold can be obtained as follow:
According to the law of conservation of energy, we have:
Heat loss = Heat gain
MC(T –Tₑ) = MᵥᵥC(Tₑ – Tᵥᵥ)
74 × C(94 – 32) = 120 × 4.184 (32 – 26.5)
C × 4588 = 2761.44
Divide both side by 4588
C = 2761.44 / 4588
C = 0.60 J/gºC
Thus, the specific heat capacity of the metal is 0.60 J/gºC
Learn more about heat transfer:
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Answer:
The compound with the correct formula is;
A. MNO₃
Explanation:
The number of oxidation states in the metal, M = One oxidation state
The formula of the compound formed by the metal, M = MHCO₃
We note that the ion HCO₃⁻, known as hydrogen carbonate has an oxidation number of -1
Similarly nitrate, NO₃⁻ has an oxidation number of -1, therefore, the metal M can form similar compound formed with HCO₃⁻ with nitrate, and we have;
The possible compounds formed by the metal 'M' includes MHCO₃ and MNO₃.
Answer:
1.76 moles of HCl.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2KMnO₄ + 8HCl —> 3Cl₂ + 2MnO₂ + 4H₂O + 2KCl
From the balanced equation above,
2 moles of KMnO₄ reacted with 8 moles of HCl.
Finally, we shall determine the number of mole of HCl needed to react with 0.44 moles of KMnO₄. This can be obtained as follow:
From the balanced equation above,
2 moles of KMnO₄ reacted with 8 moles of HCl.
Therefore, 0.44 moles of KMnO₄ will react with = (0.44 × 8)/2 = 1.76 moles of HCl.
Thus, 1.76 moles of HCl is needed for the reaction.
