What would the cathode be in a nickel and copper electrolytic cell

Carlos is trying to determine how music affects the growth of bean plants. He hypothesizes that classical music will cause a pla

lord [1]

Answer:

well what I think is that C is the correct answer

4 0

3 years ago

Suppose that on a dry, sunny day when the air temperature is near 37 ∘C,37 ∘C, a certain swimming pool would increase in tempera

ioda

Answer:

The fraction of water body necessary to keep the temperature constant is 0,0051.

Explanation:

Heat:

$Q=m*Ce*ΔT$

Q= heat (unknown)

m= mass (unknown)

Ce= especific heat (1 cal/g*°C)

ΔT= variation of temperature (2.75 °C)

Latent heat:

$ΔE=∝mΔHvap$

ΔE= latent heat

m= mass (unknown)

∝= mass fraction (unknown)

ΔHvap= enthalpy of vaporization (539.4 cal/g)

Since Q and E are equal, we can match both equations:

$m*Ce*ΔT=∝*m*ΔHvap$

Mass fraction is:

$∝=\frac{Ce*ΔT}{ΔHvap}$

$∝=\frac{(1 cal/g*°C)*2.75°C}{539.4 cal/g}$

∝=0,0051

7 0

3 years ago

What is the percent by mass of oxygen in ca(oh)2? [formula mass = 74.1]

andre [41]

16(2)/74.1=.431

.431x100= 43.1%

8 0

3 years ago

Read 2 more answers

Valproic acid, used to treat seizures and bipolar disorder, is composed of C, H, and O. A 0.165-g sample is combusted to produce

sergeinik [125]

Answer:

The empirical formula is = $C_4H_8O$

The formula of Valproic acid = $C_8H_{16}O_2$

Explanation:

Mass of water obtained = 0.166 g

Molar mass of water = 18 g/mol

Moles of $H_2O$ = 0.166 g /18 g/mol = 0.00922 moles

2 moles of hydrogen atoms are present in 1 mole of water. So,

Moles of H = 2 x 0.00922 = 0.01844 moles

Molar mass of H atom = 1.008 g/mol

Mass of H in molecule = 0.01844 x 1.008 = 0.018588 g

Mass of carbon dioxide obtained = 0.403 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of $CO_2$ = 0.403 g /44.01 g/mol = 0.009157 moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

Moles of C = 0.009157 moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 0.009157 x 12.0107 = 0.11 g

Given that the Valproic acid only contains hydrogen, oxygen and carbon. So,

Mass of O in the sample = Total mass - Mass of C - Mass of H

Mass of the sample = 0.165 g

Mass of O in sample = 0.165 - 0.11 - 0.018588 = 0.036412 g

Molar mass of O = 15.999 g/mol

Moles of O = 0.036412 / 15.999 = 0.002276 moles

Taking the simplest ratio for H, O and C as:

0.01844 : 0.002276 : 0.009157

= 8 : 1 : 4

The empirical formula is = $C_4H_8O$

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 = 4×12 + 8×1 + 16= 72 g/mol

Molar mass = 144 g/mol

So,

Molecular mass = n × Empirical mass

144 = n × 72

⇒ n = 2

The formula of Valproic acid = $C_8H_{16}O_2$

7 0

3 years ago

An air stream enters a variable area channel at a velocity of 30 m/s with a pressure of 120 kPa and a temperature of 10°C. At a

nekit [7.7K]

Answer:

P₂= 74 kPa under constant density and ρ₂ = 1.06 kg/m³ (-38.6% of error compared with incompresible assumption) . Thus Bernoulli’s equation should not be applied

Explanation:

Assuming ideal gas behaviour of air , then

P*V= n*R*T = m / M * R *T

since

ρ= m/V = P*M /( R *T)

where

n= moles , V= volume , m= mass

ρ= density

P= pressure = 120 kPa= 120000 Pa

M= molecular weight of air = 0.21*32+0.79*28= 28.24 gr/mol = 0.02824 kg/mol

T= absolute temperature = 10°C + 273 = 283 K

R= ideal gas constant = 8.314 J/mol K

solving for ρ

ρ= P*M /( R *T) = 120000 Pa*0.02824 kg/mol/(8.314 J/mol K*283 K) = 1.47 kg/m³

then from Bernoulli's equation

P₁ + ρ*v₁²/2 = P₂ + ρ*v₂²/2

where 1 denotes inlet and 2 denotes other point , p = pressure and v= velocity . Then solving for p₂

P₁ + ρ*v₁²/2 = P₂ +ρ*P₂²/2

P₂= P₁ +ρ*v₁²/2 - ρ*v₂²/2 = P₁ +ρ/2*(v₁² - v₂²)

replacing values

P₂= P₁ +ρ/2*(v₁² - v₂²) = 120000 Pa + 1.47 kg/m³/2*[(30 m/s)²-(250 m/s)²] = 74724 Pa = 74 kPa

P₂= 74 kPa

then if the temperature remains constant

ρ₁= P₁*M /( R *T) and ρ₂= P₂*M /( R *T)

dividing both equations

ρ₂/ρ₁ = P₂/ P₁

ρ₂ = (P₂/ P₁)*ρ₁

then from Bernoulli's equation

P₁ + ρ₁*v₁²/2 = P₂ + ρ₂*v₂²/2

P₂ = P₁ + ρ₁*v₁²/2 - ρ₂*v₂²/2

therefore

ρ₂ = (P₂/ P₁)*ρ₁ = (P₁ + ρ₁*v₁²/2 - ρ₂*v₂²/2 ) /P₁ *ρ₁

P₁ * ρ₂ = P₁ *ρ₁ + ρ₁²*v₁²/2 - ρ₂*ρ₁ * v₂²/2

P₁ * ρ₂ + ρ₂*ρ₁ * v₂²/2 = P₁ *ρ₁ + ρ₁²*v₁²/2

ρ₂* (P₁ + ρ₁ * v₂²/2) = P₁ *ρ₁ + ρ₁²*v₁²/2

ρ₂ = (P₁ *ρ₁ + ρ₁²*v₁²/2)/(P₁ + ρ₁ * v₂²/2) = (P₁ + ρ₁*v₁²/2)/(P₁/ρ₁ + v₂²/2)

replacing values

ρ₂ = ( 120000 Pa + 1.47 kg/m³/2*(30 m/s)²)/(120000 Pa/1.47 kg/m³+1/2*(250 m/s)²)

ρ₂ = 1.06 kg/m³

the error of assuming constant ρ would be

e = (ρ₂ - ρ)/ρ₂= 1- ρ/ρ₂= 1- 1.47 kg/m³/1.06 kg/m³ = -0.386 (-38.6%)

thus Bernoulli’s equation should not be applied

8 0

3 years ago