Ask question

Ask question

All categories

2 years ago

13

The introduction of chemical, physical, and/or biological agents into water that degrade the water quality is known as geo-pollu

tion.
a. true
b. false

1 answer:

Tatiana [17]2 years ago

8 0

A. yes it is truue......

You might be interested in

How much heat is released as the temperature of 25.2 grams of iron is decreased from 72.1°C to 9.8°C? The specific heat of iron

prisoha [69]

Answer:

$Q=-697.06\ J$

Negative sign says that release of heat.

Explanation:

The expression for the calculation of the heat released or absorbed of a process is shown below as:-

$Q=m\times C\times \Delta T$

Where,

$\Delta H$ is the heat released or absorbed

m is the mass

C is the specific heat capacity

$\Delta T$ is the temperature change

Thus, given that:-

Mass = 25.2 g

Specific heat = 0.444 J/g°C

$\Delta T=9.8-72.1\ ^0C=-62.3\ ^0C$

So,

$Q=25.2\times 0.444\times -62.3\ J=-697.06\ J$

Negative sign says that release of heat.

8 0

2 years ago

The nitrogen content of organic compounds can be determined by the Dumas method. The compound in question is first reacted by pa

MAVERICK [17]

Answer:

Mass percent of nitrogen in the compound is 13,3%

Explanation:

Dumas method is an analytical method to determine nitrogen content in samples, thus:

CₐHₓNₙ + (2a+x/2) CuO → aCO₂ + ˣ/₂ H₂O + ⁿ/₂ N₂ + (2a+x/2) Cu

As the CO₂ is removed with KOH, in the mixture you have H₂O and N₂

At 25°C. the vapor pressure of water is 23,8 torr, that means that the pressure due to N₂ is:

726torr - 23,8torr = 702,2 torr.

Using gas law:

n = PV/RT

Where:

P is pressure (702,2torr≡ 0,924 atm)

V is volume (0,0318L)

R is gas constant (0,082atmL/molK)

And T is temperature (25°C≡298,15K)

Replacing, number of moles of N₂, n, are:

n = 1,20x10⁻³moles of N₂. In grams:

1,20x10⁻³moles of N₂× $\frac{28g}{1mol}$ =<em> 0,0336 g of N₂</em>.

Thus, mass percent of nitrogen in the compound is:

$\frac{0,0336g N}{0,253gSample}$ ×100= <em>13,3%</em>

<em></em>

I hope it helps!

7 0

2 years ago

What is the balanced chemical equation for the reaction used to calculate ΔH∘f of SrCO3(s)?

Igoryamba

Answer :

Standard enthalpy of formation : It is defined as the enthalpy change for the reaction that forms one mole of compound from its elements. All the substances in their standard states.

The balanced chemical equation for the reaction is,

$Sr(s)+C(s)+\frac{3}{2}O_2(g)\rightarrow SrCO_3(s)$

In the balance reaction, Strontium (Sr), Carbon (C) and Strontium carbonate $SrCO_3$ are in solid state and oxygen is in gaseous state.

5 0

2 years ago

Lars observes a substance to be a solid and to float in water at room temperature (23°C). Based on the given properties, which s

Triss [41]

Answer:

D. Sulfur Hexafluride

Explanation:

above it says the substance floats above water at room temperature and lists some substances and their density at room temp!
we know that the density of water is 1.0 so the substance in order for it to float has to be less than 1.0 and the densities for Sulfer Hexa, are all less than 1!!

I hope this helped !!

4 0

2 years ago

4.82 g of an unknown metal is heated to 115.0∘C and then placed in 35 mL of water at 28.7∘C, which then heats up to 34.5∘C. What

nikitadnepr [17]

<u>Answer:</u> The specific heat of metal is 2.34 J/g°C

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 35 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{35mL}\\\\\text{Mass of water}=(1g/mL\times 35mL)=35g$

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of metal = 4.82 g

$m_2$ = mass of water = 35 g

$T_{final}$ = final temperature = 34.5°C

$T_1$ = initial temperature of metal = 115°C

$T_2$ = initial temperature of water = 28.7°C

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$4.82\times c_1\times (34.5-110)=-[35\times 4.186\times (34.5-28.7)]$

$c_1=2.34J/g^oC$

Hence, the specific heat of metal is 2.34 J/g°C

4 0

2 years ago