What is the volumetric flow rate in L/s of a stream of air (density = 1 kg/m3) at 1 kg/s?

Answers

Answer 1

Answer:

Answer:

Volumetric flow rate: Q = 1000 L/s

Explanation:

Volumetric flow rate, also called the rate of fluid flow, is described as volume of fluid that passes a particular point per unit time. The SI unit of volumetric flow rate is m³/s.

Whereas, mass flow rate is defined as the mass of substance that passes through a point per unit of time. SI unit is kg/s.

Given- mass flow rate: ṁ = 1 kg/s and density: ρ = 1 kg/m³

Therefore, volumetric flow rate can be calculated by

$Q = \frac{\dot{m}}{\rho } = (1 kg/s)/(1 kg/m^(3)) = 1 m^(3)/s$

Since, 1 m³/s = 1000 L/s

Therefore, volumetric flow rate: Q = 1 m³/s = 1000 L/s

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MEASUREMENT AND MATTER Interconverting temperatures in Celsius and Kelvins The metal osmium becomes superconducting at temperatures below 655.mk Calculate the temperature at which osmium becomes superconducting in degrees Celsius. Be sure your answer has the correct number of significant digits. Ac

Answers

Answer: $-272^0C$ .

Explanation:

Temperature of the gas is defined as the degree of hotness or coldness of a body. It is expressed in units like $^0C$ and $K$

These units of temperature are inter convertible.

We are given:

Temperature of the gas = $655mK=0.655K$ (1mK=0.001 K)

Converting this unit of temperature into $^0C$ by using conversion factor:

$(t-273.15)^0C=tK$

$273.15K=0^0C$

Thus $0.655K=(0.655-273.15)^0C=-272^0C$

Thus the temperature is $-272^0C$ .

6CO2 + 6H20 --> C6H12O6 + 602What is the total number of moles of CO2 needed to make 2 moles of CH1206?

Answers

Answer:

12 mol CO₂

General Formulas and Concepts:

Atomic Structure

Compounds
Moles
Mole Ratio

Stoichiometry

Analyzing reactions rxn
Using Dimensional Analysis

Explanation:

Step 1: Define

Identify

[rxn] 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

[Given] 2 mol C₆H₁₂O₆

[Solve] mol CO₂

Step 2: Identify Conversions

[rxn] 6CO₂ → C₆H₁₂O₆

Step 3: Convert

[DA] Set up: $\displaystyle 2 \ mol \ C_6H_(12)O_6((6 \ mol \ CO_2)/(1 \ mol \ C_6H_(12)O_6))$
[DA] Multiply [Cancel out units]: $\displaystyle 12 \ mol \ CO_2$

Three different samples were weighed using a different type of balance for each sample. The three were found to have masses of 0.6160959 kg, 3.225 mg, and 5480.7 g. The total mass of the samples should be reported as?

Answers

Answer:

6.1 kg

Explanation:

To obtain the total mass of the sample, we must first express each mass of the sample in the same unit of measurement.

Since the SI unit of mass is kilogram (kg), we shall express the total mass of the samples in kilogram (kg).

This is illustrated below:

Mass of the samples are:

M1 = 0.6160959 kg

M2 = 3.225 mg

M3 = 5480.7 g.

Conversion of 3.225 mg to kg

1 mg = 1×10¯⁶ kg

Therefore,

3.225 mg = 3.225 × 1×10¯⁶

3.225 mg = 3.225×10¯⁶ kg

Conversion of 5480.7 g to kg

1000 g = 1 kg

Therefore,

5480.7 g = 5480.7 /1000

5480.7 g = 5.4807 kg

Thus, we can obtain the total mass of the samples as follow:

M1 = 0.6160959 kg

M2 = 3.225×10¯⁶ kg

M3 = 5.4807 kg

Total mass =?

Total mass = M1 + M2 + M3

Total mass = 0.6160959 + 3.225×10¯⁶ + 5.4807

Total mass = 6.096799125 ≈ 6.1 kg

Therefore, the total mass of the samples is approximately 6.1 kg.

A proton is a very tiny particle inside an atom. The mass of a proton isestimated to be 0.00000000000000000000000167 g. What is the mass of a
proton in scientific notation?

Answers

Answer:

1.67 ×10^-24g

Explanation:

counting from the first digit after the point till I got to the first non zero digit

What is the value for the kinetic energyfor a n = 2 Bohr orbit electron in Joules?

Answers

Answer:

K.E. = 5.4362 × 10⁻¹⁹ J

Explanation:

The expression for Bohr velocity is:

$v=(Ze^2)/(2 \epsilon_0* n* h)$

Applying values for hydrogen atom,

Z = 1

Mass of the electron ( $m_e$ ) is 9.1093×10⁻³¹ kg

Charge of electron (e) is 1.60217662 × 10⁻¹⁹ C

$\epsilon_0$ = 8.854×10⁻¹² C² N⁻¹ m⁻²

h is Plank's constant having value = 6.626×10⁻³⁴ m² kg / s

We get that:

$v=\frac {2.185* 10^6}{n}\ m/s$

Given, n = 2

So,

$v=\frac {2.185* 10^6}{2}\ m/s$

$v=1.0925* 10^6\ m/s$

Kinetic energy is:

$K.E.=\frac {1}{2}* mv^2$

So,

$K.E.=\frac {1}{2}* 9.1093* 10^(-31)* ({1.0925* 10^6})^2$

K.E. = 5.4362 × 10⁻¹⁹ J

Assuming 100% dissociation, which of the following compounds is listed incorrectly with its van't Hoff factor i? Al2(SO4)3, i = 4 NH4NO3, i = 2 Mg(NO3)2, i = 3 Na2SO4, i = 3 Sucrose, i = 1

Answers

Answer:

- Aluminium sulfate Al2(SO4)3 dissociates in two aluminium ions and three sulfate ions, therefore, van't Hoff factor is 5 (incorrect).

Explanation:

Hello,

In this case, since the van't Hoff factor is related with the species that result from the ionization of a chemical compound, we can see that that

- Aluminium sulfate Al2(SO4)3 dissociates in two aluminium ions and three sulfate ions, therefore, van't Hoff factor is 5 (incorrect).

- Ammonium nitrate NH4NO3 dissociates in one ammonium ions and one nitrate ion, therefore, van't Hoff factor is 2 (correct).

- Sodium sulfate Na2SO4 dissociates in two sodium ions and one sulfate, therefore, van't Hoff factor is 3 (correct).

- Sucrose is not ionized, therefore, van't Hoff factor is 1 (correct).

Best regards.

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