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3 years ago

3.5 grams is equivalent to how many kilograms? 3500 kg 35,000 kg 0.35 kg 0.0035 kg

Chemistry

1 answer:

bagirrra123 [75]3 years ago

4 0

There are 1000 grams in a kg.
To convert g to kg, dovide by 1000.

3.5/1000= 0.0035 kg

Final answer: D

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Pls help

Dmitry_Shevchenko [17]

Answer:

The answer would be beryllium

You were probably confused but the correct answer would actually be a beryllium ion.

4 0

2 years ago

A sample of hydrogen gas at a pressure of 0.926 atm and a temperature of 29.5 C, occupies a volume of 457 mL. If the gas is allo

klio [65]

Answer:

V₂ = 946.72 mL

Explanation:

Given data;

Initial pressure = 0.926 atm

Initial volume = 457 mL

Temperature = constant = 29.5°C

Final pressure = 0.447 atm

Final volume = ?

Solution:

The given problem will be solved through the Boyle's law,

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume

by putting values,

P₁V₁ = P₂V₂

0.926 atm × 457 mL = 0.447 atm × V₂

V₂ = 423.18 atm. mL/ 0.447 atm

V₂ = 946.72 mL

4 0

2 years ago

Be sure to answer all parts. In the average adult male, the residual volume (RV) of the lungs, the volume of air remaining after

Alenkinab [10]

Answer:

For a: The number of moles of air present in the RV is 0.047 moles

For b: The number of molecules of gas is $2.83\times 10^{22}$

Explanation:

For a:

To calculate the number of moles, we use the equation given by ideal gas follows:

$PV=nRT$

where,

P = pressure of the air = 1.00 atm

V = Volume of the air = 1200 mL = 1.2 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of the air = $37^oC=[37+273]K=310K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of air = ?

Putting values in above equation, we get:

$1.00atm\times 1.2L=n_{air}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 310K\\n_{air}=\frac{1.00\times 1.2}{0.0821\times 310}=0.047mol$

Hence, the number of moles of air present in the RV is 0.047 moles

For b:

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.047 moles of air will contain $(0.047\times 6.022\times 10^{23})=2.83\times 10^{22}$ number of gas molecules.

Hence, the number of molecules of gas is $2.83\times 10^{22}$

7 0

3 years ago

The temperature is -14°C the air pressure in an automobile tire is 149K PA if the volume does not change what is the pressure af

Jet001 [13]

Answer: The pressure after the tire is heated to 17.3°C is 167 kPa

Explanation:

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=149kPa\\T_1=-14^0C=(273-14)=259K\\P_2=?=27.5psi\\T_2=17.3^0C=(273+17.3)=290.3K$

Putting values in above equation, we get:

$\frac{149}{259}=\frac{P_2}{290.3}\\\\P_2=167kPa$

Hence, the pressure after the tire is heated to 17.3°C is 167 kPa

4 0

3 years ago

Find the ph of a 0.250 m solution of nac2h3o2. (the ka value of hc2h3o2 is 1.80×10−5). express your answer numerically to four s

Slav-nsk [51]

Answer is: pH value of solution of NaC₂H₃O₂ is 9.07.
Chemical reaction: C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻.
Ka(HC₂H₃O₂) = 1,8·10⁻⁵.
Ka · Kb = Kw.
1,8·10⁻⁵ mol/dm³ · Kb = 1·10⁻¹⁴ mol²/dm⁶; the ionic product of water at 25°C.
Kb(C₂H₃O₂⁻) = 1·10⁻¹⁴ mol²/dm⁶ ÷ 1,8·10⁻⁵ mol/dm³.
Kb(C₂H₃O₂⁻) = 5,56·10⁻¹⁰ mol/dm³.
c(C₂H₃O₂⁻) = 0,25 M.
[OH⁻] = [HC₂H₃O₂] = x.
[C₂H₃O₂⁻] = 0,25 M - x.
Kb = [OH⁻] · [HC₂H₃O₂] / [C₂H₃O₂⁻].
5,56·10⁻¹⁰ = x² / (0,25 M -x).
Solve quadratic equation: x = [OH⁻] = 0,0000118 M.
pOH = -log[OH⁻] = -log(0,0000118M) = 4,93.
pH + pOH = 14.
pH = 14 - 4,93 = 9,07.

6 0

3 years ago