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tatuchka [14]
3 years ago
15

Difference between water bath and incubator

Chemistry
1 answer:
Nikitich [7]3 years ago
7 0

Answer:

One is to reach certain temperature, the other is to maintain the body temperature.

Explanation:

Since both are to keep warm some reactions of substances, the water bath can have temperatures since 25 ºC until 70 or 80 ºC, in the case of the incubator, the main function of this aparatus is to simulate the body temperature of a living thing. Us as humans we need a body temperature of 36 - 37ºC.

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A common radio wavelength observed coming from astronomical objects is 21 cm. What temperature is associated with this radiation
qaws [65]

Answer:

The temperature associated with this radiation is 0.014K.

Explanation:

If we assume that the astronomical object behaves as a black body, the relation between its <em>wavelength</em> and <em>temperature</em> is given by Wien's displacement law.

\lambda_{max}=\frac{b}{T}

where,

λmax is the wavelength at the peak of emission

b is Wien's displacement constant (2.89×10⁻³ m⋅K)

T is the absolute temperature

For a wavelength of 21 cm,

T=\frac{b}{\lambda _{max} } = \frac{2.89 \times 10^{-3} m.K  }{0.21m} =0.014K

8 0
3 years ago
Number 1 please fast answer
Natasha2012 [34]

Answer:

I will give you 5

Explanation:

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3 0
2 years ago
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How many mL of .450 M HCL is needed to neutralize 30mL of .150 M Ba(OH)2
Aleks04 [339]
You need .556M HCL to neutralize that
7 0
2 years ago
Analysis of a gaseous chlorofluorocarbon, CClxFy, shows that it contains 11.79% C and 69.57% Cl. In another experiment, you find
uranmaximum [27]

Answer:

The molecular formula = C_2Cl_{4}F_2

Explanation:

Moles =\frac {Given\ mass}{Molar\ mass}

% of C = 11.79

Molar mass of C = 12.0107 g/mol

<u>% moles of C = 11.79 / 12.0107 = 0.9816</u>

% of Cl = 69.57

Molar mass of Cl = 35.453 g/mol

<u>% moles of Cl = 69.57 / 35.453 = 1.9623</u>

Given that the gaseous chlorofluorocarbon only contains chlorine, flourine and carbon. So,

% of F = 100% - % of C - % of C = 100 - 11.79 - 69.57 = 18.64

Molar mass of F = 18.998 g/mol

<u>% moles of F = 18.64 / 18.998 = 0.9812</u>

Taking the simplest ratio for C, Cl and F as:

0.9816 : 1.9623 : 0.9812

= 1 : 2 : 1

The empirical formula is = CCl_2F

Also, Given that:

Pressure = 21.3 mm Hg

Also, P (mm Hg) = P (atm) / 760

Pressure = 21.3 / 760 = 0.02803 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K  

Volume = 458 mL  = 0.458 L (1 mL = 0.001 L)

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.02803 atm × 0.458 L = n × 0.0821 L.atm/K.mol × 298.15 K  

⇒n = 0.00052445 moles

Given that :  

Amount  = 0.107 g  

Molar mass = ?

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.00052445= \frac{0.107\ g}{Molar\ mass}

Molar\ mass= 204.0233\ g/mol

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 1×12.0107 + 2×35.453 + 1×18.998 = 101.9147 g/mol

Molar mass = 204.0233 g/mol

So,  

Molecular mass = n × Empirical mass

204.0233 = n × 101.9147

⇒ n = 2

<u>The molecular formula = C_2Cl_{4}F_2</u>

6 0
3 years ago
The volume of a sample gas, initially at 25 C and 158 mL, increased to 450 mL. What is the final temperature of the sample of ga
Rashid [163]

Answer:

Final temperature of the gas is  576 ^{0}\textrm{C}.

Explanation:

As the amount of gas and pressure of the gas remains constant therefore in accordance with Charles's law:

                                       \frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}

where V_{1} and V_{2} are volume of gas at T_{1} and T_{2} temperature (in kelvin scale) respectively.

Here V_{1}=158mL , T_{1}=(273+25)K=298K and V_{2}=450mL

So  T_{2}=\frac{V_{2}T_{1}}{V_{1}}=\frac{(450mL)\times (298K)}{(158mL)}=849K 

849 K = (849-273) ^{0}\textrm{C} = 576 ^{0}\textrm{C}

So final temperature of the gas is  576 ^{0}\textrm{C}.

3 0
3 years ago
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