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4 years ago

A piece of material with a mass of 755 g is heated to 84.5 c and added to 50 g of water at 5

1 answer:

6 0

Alright sorry you're getting the answer hours later, but i can help with this.
so you're looking for specific heat, the equation for it is <span>macaΔTa = - mbcbΔTb with object a and object b. that's mass of a times specific heat of a times final minus initial temperature of a equals -(mass of b times specific heat of b times final minus initial temperature of b)
</span>so putting in your values is, 755g * ca * (75 celsius - 84.5 celsius) = -(50g * cb * (75 celsius - 5 celsius))
well we know the specific heat of water is always 4180J/kg celsius, so put that in for cb
with a bit of simplification to the equation by doing everything on each side first you have, -7172.5 * ca = -14630000
divide both sides by -7172.5 so you can single out ca and you get, ca= 2039.74
add units for specific heat which are J/kg celsius and the specific heat of the material is 2039.74 J/kg celsius

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2 years ago

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Answer:

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Explanation:

This is a problem in calorimetry — the measurement of the quantities of heat that flow from one object to another.

It is based on the Law of Conservation of Energy — Energy can be transformed from one type to another, but it cannot be destroyed or created.

If heat flows out of the reactor (negative), the same amount of heat must flow into the water (positive).

Since there is no change in total energy,

heat₁ + heat₂ = 0

The symbol for the quantity of heat transferred is q, so we can rewrite the word equation as

q₁ + q₂ = 0

The formula for the heat absorbed or released by an object is

q = mCΔT, where

m = the mass of the sample

C = the specific heat capacity of the sample, and

ΔT = T_f - T_i = the change in temperature

1. Equation

There are two heat flows in this problem,

heat released by reactor + heat absorbed by water = 0

q₁ + q₂ = 0

q₁ + m₂C₂ΔT₂ = 0

2. Data:

q₁ = -23 746 kJ

m₂ = ?; C₂ = 4.184 J°C⁻¹g⁻¹; T_f = 95 °C; T_i = 10 °C

3. Calculations

(a) Convert kilojoules to joules

$q_{1} = -\text{23 746 kJ} \times \dfrac{\text{1000 J}}{\text{1 kJ}} = -\text{23 746 000 J}$

(b) ΔT

ΔT₂ = T_f - T_i = 95 °C - 10 °C = 85 °C

(c) m₂

$\begin{array}{rcl}q_{1} + q_{2} & = & 0\\\text{-23 746 000 J} + m_{2} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 85 \, ^{\circ}\text{C} & = & 0\\\text{-23 746 000 J} + 356m_{2} \text{J$\cdot$g}^{-1} & = & 0\\356m_{2} \text{g}^{-1} & = & 23746000\\m_2&=& \dfrac{23746000}{\text{356 g}^{-1}}\\\\ & = & \textbf{67000 g}\\\end{array}\\$

$\text{You must circulate $\large \boxed{\textbf{67 000 g}}$ of water each hour.}$

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3 years ago

The pipet used for transferring the pickle juice to the flask was dirty and consequentially some of the pickle juice stuck to th

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Explanation:

Since, some of the given sample is stuck inside and behind the pipet. Hence, there will occur a decrease in the percent of acetic acid.

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For example, we took 10 mg of a pickel sample but 3 mg of the sample remain stuck in the pipet. This means we actually titrating a sample less than 10 mg.

Therefore, the analyte concentration in the pickel will also be less.

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4 years ago

[ 1.3 ] What is the concentration (%) of solute in the RBC (i.e., what is the concentration of all solute molecules combined?) A

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Answer:

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3 years ago

What is the molecular formula for C5H7 if its molar mass is 536 g/mol?

hodyreva [135]

Answer:

(C5H7)8

Explanation:

It's empirical formula is given as; C5H7

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To find the molecular formula;

Molecular formula = n × empirical formula

Thus;

n = 536/((12 × 5) + (1 × 7))

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Thus;

Molecular formula = (C5H7)8

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3 years ago