Which of the following body system does the immune system rely on to provide the first line of defense against pathogens

How many sig figs 18.01528

Ulleksa [173]

Answer:7

Explanation:

5 0

3 years ago

What is the symbol for temperature? m q T J

Tcecarenko [31]

Answer:

T

Explanation:

m is mass

q is heat

J is energy

6 0

2 years ago

Which of these observations helped Rutherford determine that atoms have tiny, dense, positively charged nuclei?

gladu [14]

The third reason helped Rutherford to discover the nucleus.

5 0

3 years ago

During a laboratory experiment, a 2.36-gram sample of NaHCO3 was thermally decomposed. In this experiment, carbon dioxide and wa

eduard

Answer:

90.7 %

Explanation:

1) Chemical equation (given)

2NaHCO₃ → Na₂CO₃ + H₂CO₃

2) Theoretical yield

a) Convert mass of NaHCO₃ to moles:

n = mass in grams / molar mass
molar mass = 84.007 g/mol
n = 2.36 g / 84.007 g/mol = 0.02809 mol

b) Mole ratio:

2 mol NaHCO₃ : 1 mol H₂CO₃

c) Proportionality:

2 mol NaHCO₃ / mol H₂CO₃ = 0.02809 mol NaHCO₃ / x

⇒ x = 0.2809 / 2 mol H₂CO₃ = 0.01405 mol H₂CO₃

3) Actual yield

a) Mass balance: 2.36 g - 1.57 g = 0.79 g

b) Convert 0.79 g of carbonic acid to number of moles:

n = mass in grams / molar mass

molar mass = 62.03 g/mol

n = 0.79 g / 62.03 g/mol = 0.01274 mol

4) Percentage yield, y (%)

y (%) = actual yield / theoretical yield × 100

y (%) = 0.1274 mol / 0.1405 mol × 100 = 90.68%

The answer must show 3 significant figures, so y(%) = 90.7%.

5 0

3 years ago

A sheet of polyethylene 1.5-mm thick is being used as an oxygen barrier at a temperature of 600K. If the flux is 2.48 x 10-5 kg/

Eddi Din [679]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration of high-pressure side is $C_1 = 0.722 \ kg/m^3$

Explanation:

From the question we are told that

The thickness of the polyethylene is $d = 1.5 \ mm = 0.0015 \ m$

The temperature is $T = 600 \ K$

The flux is $JA = 2.48 *10^{-5} \ kg/m^2\cdot s$

The concentration on the low-pressure side is $C_2 = 0.5 \ kg/m^3$

The initial diffusivity is $D_o = 6.2 *10^{-4} \ m^2 /s$

The activation energy for diffusion is $Q_d = 41 \ kJ /mol = 41*10^3 J /mol$

Generally the diffusivity of the oxygen at 600 K can be mathematically evaluated as

$D = D_o * e^{- \frac{Q_d}{R * T } }$

substituting values

$D = 6.2*10^{-4} * e^{- \frac{41 *10^3}{8.314 * 600 } }$

$D = 1.671 *10^{-7} \ m^2 /s$

Generally the flux is mathematically represented as

$JA = D * \frac{C_1 -C_2}{d}$

Where $C_1$ is the concentration of oxygen at the higher side

So

$C_1 = d * \frac{JA}{D} + C_2$

substituting values

$C_1 = 0.0015 * \frac{2.48*10^{-5}}{1.671*10^{-7}} + 0.5$

$C_1 = 0.722 \ kg/m^3$

6 0

3 years ago