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Alborosie
3 years ago
11

What is a hygrometer?

Chemistry
1 answer:
Archy [21]3 years ago
3 0
A hygrometer is an instrument for measuring the humidity of the air or a gas .
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What is the name of the term that describes the conditions that are kept the same between all groups in an experiment?
Colt1911 [192]
Controlled variable
6 0
3 years ago
Two objects are brought into contact Object 1 has mass 0.76 kg, specific heat capacity 0.87) g'c and initial temperature 52.2 'C
taurus [48]

Answer:

T_F=77.4\°C

Explanation:

Hello there!

In this case, according to the given information, it turns out possible to set up the following energy equation for both objects 1 and 2:

Q_1=-Q_2

In terms of mass, specific heat and temperature change is:

m_1C_1(T_F-T_1)=-m_2C_2(T_F-T_2)

Now, solve for the final temperature, as follows:

T_F=\frac{m_1C_1T_1+m_2C_2T_2}{m_1C_1+m_2C_2}

Then, plug in the masses, specific heat and temperatures to obtain:

T_F=\frac{760g*0.87\frac{J}{g\°C} *52.2\°C+70.7g*3.071\frac{J}{g\°C}*154\°C}{760g*0.87\frac{J}{g\°C} +70.7g*3.071\frac{J}{g\°C}} \\\\T_F=77.4\°C

Yet, the values do not seem to have been given correctly in the problem, so it'll be convenient for you to recheck them.

Regards!

4 0
3 years ago
If the sun is next to the earth on the left, and the moon is next to the earth on the right, what type of tide is it?
Andrej [43]

Answer:

When the sun, moon, and Earth are in alignment (at the time of the new or full moon), the solar tide has an additive effect on the lunar tide, creating extra-high high tides, and very low, low tides—both commonly called spring tides.

3 0
2 years ago
The concentration of pb2+ in a solution saturated with pbbr2(s) is 2.14 ✕ 10-2 m. calculate ksp for pbbr2.
Ugo [173]
Concentration = 2.14 âś• 10-2 m 
For [Br-], there are 2 ions so 2 x 2.14 x 10^-2 =4.28 x 10^-2  
Ksp = [Pb][Br]^2 = 2.14 âś• 10-2 x (4.28 x 10^-2 )^2 = 39.20 x 10^-6 
Ksp = 3.92 x 10^-5
3 0
3 years ago
Read 2 more answers
What is the quantity of heat (in kJ) associated with cooling 185.5 g of water from 25.60°C to ice at -10.70°C?Heat Capacity of S
Cerrena [4.2K]

Taking into account the definition of calorimetry, sensible heat and latent heat,  the amount of heat required is 37.88 kJ.

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

<h3>Sensible heat</h3>

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

<h3>Latent heat</h3>

Latent heat is defined as the energy required by a quantity of substance to change state.

When this change consists of changing from a solid to a liquid phase, it is called heat of fusion and when the change occurs from a liquid to a gaseous state, it is called heat of vaporization.

  • <u><em>25.60 °C to 0 °C</em></u>

First of all, you should know that the freezing point of water is 0°C. That is, at 0°C, water freezes and turns into ice.

So, you must lower the temperature from 25.60°C (in liquid state) to 0°C, in order to supply heat without changing state (sensible heat).

The amount of heat a body receives or transmits is determined by:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

  • c= Heat Capacity of Liquid= 4.184 \frac{J}{gC}
  • m= 185.5 g
  • ΔT= Tfinal - Tinitial= 0 °C - 25.60 °C= - 25.6 °C

Replacing:

Q1= 4.184 \frac{J}{gC}× 185.5 g× (- 25.6 °C)

Solving:

<u><em>Q1= -19,868.98 J</em></u>

  • <u><em>Change of state</em></u>

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to

Q = m×L

where L is called the latent heat of the substance and depends on the type of phase change.

In this case, you know:

n= 185.5 grams× \frac{1mol}{18 grams}= 10.30 moles, where 18 \frac{g}{mol} is the molar mass of water, that is, the amount of mass that a substance contains in one mole.

ΔHfus= 6.01 \frac{kJ}{mol}

Replacing:

Q2= 10.30 moles×6.01 \frac{kJ}{mol}

Solving:

<u><em>Q2=61.903 kJ= 61,903 J</em></u>

  • <u><em>0 °C to -10.70 °C</em></u>

Similar to sensible heat previously calculated, you know:

  • c = Heat Capacity of Solid = 2.092 \frac{J}{gC}
  • m= 185.5 g
  • ΔT= Tfinal - Tinitial= -10.70 °C - 0 °C= -10.70 °C

Replacing:

Q3= 2.092 \frac{J}{gC} × 185.5 g× (-10.70) °C

Solving:

<u><em>Q3= -4,152.3062 J</em></u>

<h3>Total heat required</h3>

The total heat required is calculated as:  

Total heat required= Q1 + Q2 +Q3

Total heat required=-19,868.98 J + 61,903 J -4,152.3062 J

<u><em>Total heat required= 37,881.7138 J= 37.8817138 kJ= 37.88 kJ</em></u>

In summary, the amount of heat required is 37.88 kJ.

Learn more about calorimetry:

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7 0
2 years ago
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