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777dan777 [17]
3 years ago
11

What is the change in temperature when 75 grams of water releases -2657 J of energy? The specific heat of water is 4.18 J/g°C

Chemistry
1 answer:
Naily [24]3 years ago
8 0
I think the answer for this is 4702.5 J/g*k Depending on if it is water as a solid liquid or gas. I used water as a liquid when I solved it. J=(75g)(4.18 J/g*k)(15K)
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Real life example of when temperature increases, then the pressure increases.
timurjin [86]

Answer:

Tyre

In cold weather, you might have regularly kept a check on the pressure of the tyres of your car. Driving increases the temperature of the tyres, and, therefore, the air inside the tyre warms and expands. When you measure the pressure of the tyres at the time when you have just driven the car, it will be high. However, in cold weather, the pressure of the tyres will be low. So, it is recommended that you should always measure the pressure of the tyres.

6 0
2 years ago
Describe two uses of surfactants. At least one must be something that was not described in the passage. For each example, explai
notka56 [123]
There are several types of surfactants, to name two: alkylbenzene sulfonates  or detergents; and lauryl sulfate or foaming agents. 

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8 0
3 years ago
Read 2 more answers
A piston chamber filled with ideal gas is kept in a constant-temperature bath at 25.0°C. The piston expands from 25.0 mL to 75.0
larisa [96]

Answer : The work done by the system is, 2.2722 J

Explanation :

The expression used for work done in reversible isothermal expansion will be,

w=nRT\ln (\frac{V_2}{V_1})

where,

w = work done = ?

n = number of moles of gas  = 0.00100 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas  = 25^oC=273+25=298K

V_1 = initial volume of gas  = 25 mL

V_2 = final volume of gas  = 75 mL

Now put all the given values in the above formula, we get:

w=0.00100mole\times 8.314J/moleK\times 298K\times \ln (\frac{75}{25})

w=2.722J

Therefore, the work done by the system is, 2.2722 J

8 0
3 years ago
47.0ml of a HBr solution were titrated with 37.5ml of a 0.215M LiOH solution to reach the equivalence point. what is the molarit
faltersainse [42]

Hello!

The molarity of the HBr solution is 0,172 M.

Why?

The neutralization reaction between LiOH and HBr is the following:

HBr(aq) + LiOH(aq) → LiBr(aq) + H₂O(l)

To solve this exercise, we are going to apply the common titration equation:

M1*V1=M2*V2

M1=\frac{M2*V2}{V1}= \frac{0,215 M * 37,5 mL}{47 mL}=0,172 M

Have a nice day!

4 0
3 years ago
Titration of 0.824 g of potassium hydrogen phthalate required 38.314 g of naoh solution to reach the end point detected by pheno
melamori03 [73]

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

3 0
3 years ago
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