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2 years ago

Describe two human activities that can affect the water cycle.

Chemistry

2 answers:

kakasveta [241]2 years ago

7 0

Water pollution and global warming

vova2212 [387]2 years ago

4 0

Deforestation....trees release water vapour..with the trees being cut down less water vapour is being released..
consuming the water.... we use up water for drinking and for our homes etc...
hope this helps!

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Analyze the following blood splatter pictures to determine the general direction, origin, and impact

kkurt [141]

Answer:

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2 years ago

a) One kmol of N2O4 dissociates at 25℃, 1 atm to form an equilibrium ideal gas mixture of N2O4 and NO2 in which the amount of N2

horsena [70]

Answer:

neq N2O4 = 0.9795 mol.....P = 0.5 atm; T = 25°C

Explanation:

ni change eq.

N2O4 1 1 - x 0.8154.....P = 1 atm; T = 25°C

NO2 0 0 + x x

∴ x = neq = Peq.V / R.T.....ideal gas mix

if P = 0.5 atm, T = 25°C; assuming: V = 1 L

⇒ x = neq = ((0.5 atm)(1 L))/((0.082 atm.L/K.mol)(298 K))

⇒ x = neq = 0.0205 mol

⇒ neq N2O4 = 1 - x = 1 - 0.0205 = 0.9795 mol

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2 years ago

What material was added to powdered rock during tuttle and bowen's experiments?

Rzqust [24]

Answer:

water was added to powdered rock

Explanation:

4 0

2 years ago

A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

sashaice [31]

Answer:

a) 90 kg

b) 68.4 kg

c) 0 kg/L

Explanation:

Mass balance:

$-w=\frac{dm}{dt}$

w is the mass flow

m is the mass of salt

$-v*C=\frac{dm}{dt}$

v is the volume flow

C is the concentration

$C=\frac{m}{V+(6-3)*L/min*t}$

$-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}$

$-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}$

$-3*L/min*\frac{dt}{2000L+(3)*L/min*t}=\frac{dm}{m}$

$-3*L/min*\int_{0}^{t}\frac{dt}{2000L+(3)*L/min*t}=\int_{90kg}^{m}\frac{dm}{m}$

$-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)$

$-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)$

$m=90kg*[2000L/(2000L+3*L/min*t)]$

a) Initially: t=0

$m=90kg*[2000L/(2000L+3*L/min*0)]=90kg$

b) t=210 min (3.5 hr)

$m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg$

c) If time trends to infinity the division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.

6 0

2 years ago

The most concentrated solution is

aev [14]

The most concentrated solution is b

6 0

2 years ago

Read 2 more answers