An empty beaker weighs 32.4257 grams. A 10 ml pipet sample of an unknown liquid is

What are the effects on the bodies of water? use terms algal bloom, dead zone and hypotoxic

Anna35 [415]

Water pollution is defined as the mixing of water with unwanted substances and that makes water unsafe.

Water pollution includes runoff of excess fertilizers,

herbicides, and insecticides from agricultural lands

and residential areas; oil, grease, and toxic chemicals

from urban runoff and energy production; and

sediment from improperly managed construction sites,

crop and forest lands, and eroding stream banks.

Polluted waters have high BOD which affects aquatic biodiversity.

Hpotoxic substances such as lead, arsenic, and fluoride cause many problems to aquatic animals and humans. Water contaminated with Arsenic results in diseases such as arsenicosis. Fluoride had been reported to cause depression in DNA and RNA synthesis in cultured cells, significant reductions in DNA and RNA levels, and conditions including aging, cancer, and arteriosclerosis are associated with DNA damage and its disrepair. Lead causes problems related to the central nervous system. Children and pregnant women are most at risk. Routine applications of fertilizers and pesticides for agriculture and uncontrolled runoff in water bodies. Adds nitrogen and phosphorus to water. It adds nitrogen and phosphorus to water causing eutrophication and algal blooms.

Therefore, all these activities make our water bodies a death zone for the marine ecosystem as well as humans.

To know more about water pollution, refer to the below link:

brainly.com/question/2976496

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8 0

2 years ago

1. The heat of fusion for the ice-water phase transition is 335 kJ/kg at 0°C and 1 bar. The density of water is 1000 kg/m3 at th

vodomira [7]

Answer:

Expression for the change of melting temperature with pressure..> T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa), Freezing Point = 0°C

Explanation:

Derivation from state postulate

Using the state postulate, take the specific entropy, , for a homogeneous substance to be a function of specific volume and temperature .

ds = (partial s/partial v)(t) dv + (partial s/partial T)(v) dT

During a phase change, the temperature is constant, so

ds = (partial s/partial v)(T) dv

Using the appropriate Maxwell relation gives

ds = (partial P/partial T)(v) dv

s(β) – s(aplαha) = dP/dT (v(β) – v(α))

dP/dT = s(β) – s(α)/v(β) – v(α) = Δs/Δv

Here Δs and Δv are respectively the change in specific entropy and specific volume from the initial phase α to the final phase β.

For a closed system undergoing an internally reversible process, the first law is

du = δq – δw = Tds - Pdv

Using the definition of specific enthalpy, h and the fact that the temperature and pressure are constant, we have

du + Pdv = dh Tds,

ds = dh/T,

Δs = Δh/T = L/T

After substitution of this result into the derivative of the pressure, one finds

dp/dT = L/TΔv

<u>This last equation is the Clapeyron equation.</u>

a)

(dP/dT) = dH/TdV => dP/dlnT = dH/dV

=> dP/dlnT = dH/dV = [H(liquid) - H(solid)]/[V(liquid) - V(solid)]

= [335,000 J/kg]/[1000⁻¹ - 915⁻¹ m³/kg]

= -3.61x10⁹ J/m³ = -3.61x10⁹ Pa

=> P₂ = P₁ - 3.61x10⁹ ln(T₂/T₁) Pa

or

T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa)

b) if the pressure in Denver is 84.6 kPa:

T₂(freezing) = 273.15exp[-(84,600-100,000)/(3.61x10⁹)]

≅ 273.15 = 0°C T₁(freezing) essentially no change

5 0

3 years ago

(06.03 MC) A 50.0 mL sample of gas at 20.0 atm of pressure is compressed to 40.0 atm of pressure at constant temperature. What i

oee [108]

Answer:

New volume is 25.0 mL

Explanation:

Let's assume the gas sample behaves ideally.

According to combined gas law for an ideal gas-

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

where, $P_{1}$ and $P_{2}$ represent initial and final pressure respectively

$V_{1}$ and $V_{2}$ represent initial and final volume respectively

$T_{1}$ and $T_{2}$ represent initial and final temperature (in kelvin) respectively

Here, $T_{1}=T_{2}$ , $V_{1}=50.0mL$ , $P_{1}=20.0atm$ and $P_{2}=40.0atm$

So, $V_{2}=\frac{P_{1}V_{1}T_{2}}{P_{2}T_{1}}=\frac{(20.0atm)\times (50.0mL)}{40.0mL}=25.0mL$

So, the new volume is 25.0 mL

6 0

3 years ago

A 0.708 g sample of a metal, M, reacts completely with sulfuric acid according to the reaction M ( s ) + H 2 SO 4 ( aq ) ⟶ MSO 4

ollegr [7]

Answer:

The metal has a molar mass of 65.37 g/mol

Explanation:

Step 1: Data given

Mass of the metal = 0.708 grams

Volume of hydrogen = 275 mL = 0.275 L

Atmospheric pressure = 1.0079 bar = 0.9947 atm

Temperature = 25°C

Vapor pressure of water at 25 °C = 0.03167 bar = 0.03126 atm

Step 2: The balanced equation

M(s) + H2SO4(aq) ⟶ MSO4 (aq) + H2(g)

Step 3: Calculate pH2

Atmospheric pressure = vapor pressure of water + pressure of H2

0.9947 atm = 0.03126 atm + pressure of H2

Pressure of H2 = 0.9947 - 0.03126

Pressure of H2 = 0.96344 atm

Step 4: Calculate moles of H2

p*V=n*R*T

⇒ with p = The pressure of H2 = 0.96344 atm

⇒ with V = the volume of H2 = 0.275 L

⇒ with n = the number of moles H2 = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 25°C = 298 Kelvin

n = (p*V)/(R*T)

n = (0.96344 * 0.275)/(0.08206*298)

n = 0.01083 moles

Step 5: Calculate moles of M

For 1 mole of H2 produced, we need 1 mole M

For 0.0108 moles of H2 we need 0.01083 moles of M

Step 6: Calculate molar mass of M

Molar mass M = Mass M / moles M

Molar mass M = 0.708 grams / 0.01083 moles

Molar mass M = 65.37 g/mol

The metal has a molar mass of 65.37 g/mol

5 0

3 years ago

Difference between ntp and stp

Phantasy [73]

Answer:

STP stands for Standard Temperature and Pressure. NTP stands for Normal Temperature and Pressure.

Explanation:

STP is set by the IUPAC as 0°C and 100 kPa or 1 bar.

NTP is set at 101.325 kPa but uses 20°C as the temperature

8 0

3 years ago