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3 years ago

How does a thick layer of algae growing on the surface of a pond affect the lives of other organisms living in the pond ?

Chemistry

2 answers:

Naily [24]3 years ago

7 0

They might not be able to get enough sunlight in some cases and might die from it

nignag [31]3 years ago

7 0

Answer:

The other organisms will die.

Explanation:

algae aborbs a lot of oxygen, too much oxygen intake suffocates the other organisms around it.

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What really is the Ebola virus?

Sloan [31]

The Ebola virus belongs to a family of viruses termed Filoviridae. Filovirus particles form long sometimes branched filaments of varying shapes, as well as shorter filaments , and may measure up to 14,000 nanometers in length with diameter of 80 nanometers. Viral particles contain one molecule of single stranded RNA enveloped in a lipid membrane. New viral particle bud from the surface of their host cell. Although Ebola virus was only discovered in 1976, it is an ancient virus and is thought to have split from other viruses thousands of years ago.

7 0

3 years ago

Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,

WITCHER [35]

Answer: Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]$

We are given:

$\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ$

Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

6 0

3 years ago

Suppose that 25.0 mL of 0.10 M CH3COOH (aq) is titrated with 0.10 M NaOH (aq). What is the pH after the addition of 10.0 mL of 0

olya-2409 [2.1K]

Answer:

$pH=-1.37$

Explanation:

We are given that 25 mL of 0.10 M $CH_3COOH$ is titrated with 0.10 M NaOH(aq).

We have to find the pH of solution

Volume of $CH_3COOH=25mL=0.025 L$

Volume of NaoH=0.01 L

Volume of solution =25 +10=35 mL= $\frac{35}{1000}=0.035 L$

Because 1 L=1000 mL

Molarity of NaOH=Concentration OH-=0.10M

Concentration of H+= Molarity of $CH_3COOH$ =0.10 M

Number of moles of H+=Molarity multiply by volume of given acid

Number of moles of H+= $0.10\times 0.025$ =0.0025 moles

Number of moles of $OH^-=0.10\times 0.01$ =0.001mole

Number of moles of H+ remaining after adding 10 mL base = 0.0025-0.001=0.0015 moles

Concentration of H+= $\frac{0.0015}{0.035}=4.28\times 10^{-2} m/L$

pH=-log [H+]=-log [4.28 $\times 10^{-2}$ ]=-log4.28+2 log 10=-0.631+2

$pH=-1.37$

6 0

3 years ago

What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.0×1015Hz?

RideAnS [48]

The kinetic energy of the emitted electrons of cesium when it is exposed to UV rays of frequency $1.0 \times {10^{15}}\;{\text{Hz}}$ is $\boxed{6.63 \times {{10}^{ - 19}}\;{\text{J}}}$

Further Explanation:

Photoelectric effect:

When light is made to fall on any substance, electrons are emitted from it. This is known as the photoelectric effect and the emitted electrons are called photoelectrons. The electrons are emitted because of the transference of energy from light to the electrons.

Cesium is a member of the alkali metal group so it is highly reactive and shows photoelectric effect to the maximum extent. It can remove its electron so easily because of its atomic size. Due to large atomic size of cesium, its outermost electrons are held very less tightly to the nucleus and therefore removed easily.

According to the Planck-Einstein equation, the energy is proportional to the frequency and is expressed as follows:

${\mathbf{E=h\nu }}$ ......(1)

Here,

${\text{E}}$ is the energy.

$h$ is the Plank’s constant.

$\nu$ is the frequency.

The frequency of UV rays is $1.0 \times {10^{15}}\;{\text{Hz}}$ or $1.0 \times {10^{15}}\;{{\text{s}}^{ - 1}}$

The value of Planck’s constant is $6.626 \times {10^{ - 34}}\;{\text{J}}\cdot{\text{s}}$ .

Substitute these values in equation (1)

$\begin{aligned}{\text{E}}&=\left( {6.626 \times {{10}^{ - 34}}\;{\text{J}}\cdot{\text{s}}}\right)\left( {1.0 \times {{10}^{15}}\;{{\text{s}}^{ - 1}}}\right)\\&=6.63\times {10^{ - 19}}\;{\text{J}}\\\end{aligned}$

But when electrons are ejected out from the surface of the substance, all of its energy is considered as kinetic energy.

So the kinetic energy of the electrons is ${\mathbf{6}}{\mathbf{.63 \times 1}}{{\mathbf{0}}^{{\mathbf{ - 19}}}}\;{\mathbf{J}}$ .

Learn more:

1. Statement about subatomic particle: brainly.com/question/3176193

2. The energy of a photon in light: brainly.com/question/7590814

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Structure of the atom

Keywords: kinetic energy, frequency, energy, photoelectric effect, Planck's constant, light, electrons, photoelectrons, proportional, transference, reactive, cesium.

7 0

3 years ago

Read 2 more answers

.7 liters of CO represents how many molecules?

Ronch [10]

Answer:

1.88 × 10²² Molecules of CO

Explanation:

At STP for an ideal gas,

Volume = Mole × 22.4 L/mol

Or,

Mole = Volume / 22.4 L/mol

Mole = 0.7 L / 22.4 L/mol

Mole = 0.03125 moles

Now,

No. of Molecules = Moles × 6.022 × 10²³ Molecules/mol

No. of Molecules = 0.03125 × 6.022 × 10²³ Molecules/mol

No. of Molecules = 1.88 × 10²² Molecules of CO

7 0

2 years ago