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steposvetlana [31]
2 years ago
7

According to the diagram below, a bacterial cell does not contain

Chemistry
1 answer:
Pavlova-9 [17]2 years ago
3 0

Answer:

nucleus

Explanation:

You might be interested in
A sample of 42 mL of carbon dioxide gas was placed in a piston in order to maintain a constant 101 kPa of pressure.
Lilit [14]

Answer:

The answer to your question is 33.4 ml

Explanation:

Data

volume 1 = V1 = 42 ml

temperature 1 = T1 = 20°C

temperature 2 = T2 = -60°C

Volume 2 = V2 = x

Process

1.- Convert celsius to kelvin

T1 = 20 + 273 = 293°K

T2 = -60 + 273 = 233°K

2.- Use the Charles' law to solve this problem

               \frac{V1}{T1} = \frac{V2}{T2}

Solve for V2

                V2 = \frac{V1T2}{T1}

3.- Substitution

               V2 = \frac{(42)(233)}{293}

4.- Simplification

                V2 = \frac{9786}{293}

5.- Result

                V2 = 33.4ml

3 0
3 years ago
A freezer compartment is covered with a 2-mm-thick layer of frost at the time it malfunctions. If the compartment is in ambient
ruslelena [56]

Answer:

The time required to melt the frost is 3.25 hours.

Explanation:

The time required to melt the frost dependes on the latent heat of the frost and the amount of heat it is transfered by convection to the air .

The heat transferred per unit area can be expressed as:

q=h_c*A*\Delta T\\\\q/A=h_c*\Delta T

being hc the convective heat transfer coefficient (2 Wm^-2K^-1) and ΔT the difference of temperature (20-0=20 °C or K).

q/A=h_c*\Delta T=2\frac{W}{m^2K}*20K=40\frac{W}{m^2}

If we take 1 m^2 of ice, with 2 mm of thickness, we have this volume

V=T*A = 0.002 m * 1 m^2=0.002m^3

The mass of the frost can be estimated as

M=\rho * V=700\frac{kg}{m^3}*0.002m^3= 1.4 kg

Then,  the amount of heat needed to melt this surface (1 m²) of frost is

Q=L*M=334\frac{kJ}{kg}*1.4kg= 467.6kJ

The time needed to melt the frost can be calculated as

t=\frac{Q}{(q/A)}=\frac{467.6kJ/m2}{40W/m2} = 11.69\frac{kJ}{W}*\frac{1W*s}{1J}*\frac{1000J}{1kJ}=   11690s=3.25h

7 0
3 years ago
What is the Relative Formula Mass of Potassium bromide?
Aleonysh [2.5K]
The relative formua for Potassium bromide is Kbr
6 0
3 years ago
A solution is made by combining 500 mL of 0.10 M HF (Ka=7.2 x 10^-4) with 300 mL of 0.15 M NaF. What is the pH of the resulting
n200080 [17]

Answer:

b) 3.10

Explanation:

HF ⇄ H + + F

Using Henderson-Hasselbalch Equation:

pH = pKa + log [A-]/[HA].

Where;

pKa = Dissociation constant = -log Ka

Hence, pKa of HF = -log 7.2 x 10^-4 = 3.14266

[A-] = concentration of conjugate base after dissociation = moles of base/total volume

          = 0.15 x 0.3/0.8

               = 0.05625 M

[HA] = concentration of the acid = moles of acid/total volume

             = 0.10 x 0.5/0.8

                    = 0.0625 M

Note: <em>Total volume = 500 + 300 = 800 mL = 0.8 dm3</em>

pH = 3.14266 + log [0.05625/0.0625]

      = 3.14267 + (-0.04575749056)

           = 3.09691250944

<em>From all the available options below:</em>

<em>a) 2.97 </em>

<em>b) 3.10 </em>

<em>c) 3.19 </em>

<em>d) 3.22 </em>

<em>e) 3.32</em>

The correct option is b.

4 0
3 years ago
15g baking soda is dissolved in 100 mL water. The solute of the
Mumz [18]

Answer:

Baking soda

Explanation:

5 0
3 years ago
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