Answer:
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Answer:
982.5 kg/m³
Explanation:
When the temperature of a fluid increases, it dilates, and because of the variation of the volume, it's density will vary too. The density can be calculated by the expression:
ρ₁ = ρ₀/(1 + β*(t₁ - t₀))
Where ρ₁ is the final density, ρ₀ the initial density, β is the constant coefficient of volume expansion, t₁ the final temperature, and t₀ the initial temperature.
At t₀ = 4°C, the water desity is ρ₀ = 1,000 kg/m³. The value of the constant for water is β = 0.0002 m³/m³ °C, so, for t₁ = 93°C
ρ₁ = 1,000/(1 + 0.0002*(93 - 4))
ρ₁ = 1,000/(1+ 0.0178)
ρ₁ = 982.5 kg/m³
B KOH
I would say this is the base for the compound substance
Given that, an experiment to measure the enthalpy change for the reaction of aqueous copper(II) sulfate, CuSO4(aq) and zinc, Zn(s) was carried out in a coffee cup calorimeter; the heat of the reaction in the whole system is calculated to be 2218.34 kJ
Heat of reaction (i.e enthalpy of reaction) is the quantity of heat that is required to be added or removed when a chemical reaction is taken place in order to maintain all of the compounds present at the same temperature.
The formula used to calculate the heat of the reaction can be expressed as follows:
Q = mcΔT
where:
- Q = quantity of heat transfer
- m = mass
- c = specific heat of water = 4.18 kJ/g °C (constant)
- ΔT = change in temparature
From the information given:
- The initial temperature (T₁) = 25° C
- The final temperature (T₂) = 91.5° C
∴
The change in temperature i.e. ΔT = T₂ - T₁
ΔT = 91.5° C - 25° C
ΔT = 66.5° C
The number of moles of CuSO₄ = 1.00 mol/dm³ × 50.0 cm³
= 0.05 moles
- Since the molar mass of CuSO₄ = 159.609 g/mol
Then;
Using the relation:
By crossing multiplying;
mass of CuSO₄ = number of moles of CuSO₄ × molar mass of CuSO₄
mass of CuSO₄ = 0.05 moles × 159.609 g/moles
mass of CuSO₄ = 7.9805 grams
∴
Using the formula from above:
Q = mcΔT
Q = 7.9805 g × 4.18 kJ/g °C × 66.5° C
Q = 2218.34 kJ
Therefore, we can conclude that the heat of the reaction is 2218.34 kJ
Learn more about the chemical reaction here:
brainly.com/question/20250226?referrer=searchResults
1.062 mol/kg.
<em>Step 1</em>. Write the balanced equation for the neutralization.
MM = 204.22 40.00
KHC8H4O4 + NaOH → KNaC8H4O4 + H2O
<em>Step 2</em>. Calculate the moles of potassium hydrogen phthalate (KHP)
Moles of KHP = 824 mg KHP × (1 mmol KHP/204.22 mg KHP)
= 4.035 mmol KHP
<em>Step 3</em>. Calculate the moles of NaOH
Moles of NaOH = 4.035 mmol KHP × (1 mmol NaOH/(1 mmol KHP)
= 4.035 mmol NaOH
<em>Step 4</em>. Calculate the mass of the NaOH
Mass of NaOH = 4.035 mmol NaOH × (40.00 mg NaOH/1 mmol NaOH)
= 161 mg NaOH
<em>Step 5</em>. Calculate the mass of the water
Mass of water = mass of solution – mass of NaOH = 38.134 g - 0.161 g
= 37.973 g
<em>Step 6</em>. Calculate the molal concentration of the NaOH
<em>b</em> = moles of NaOH/kg of water = 0.040 35 mol/0.037 973 kg = 1.062 mol/kg
