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3 years ago

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Ancient cyanobacteria released ______, which assisted in creating the atmosphere as we know it today.

2 answers:

zavuch27 [327]3 years ago

8 0

Ancient cyanobacteria released <span>oxygen
</span>which assisted in creating the atmosphere as we know it today.

andrey2020 [161]3 years ago

3 0

Answer;

- Oxygen

-Ancient cyanobacteria released oxygen, which assisted in creating the atmosphere as we know it today.

Explanation;

-The oxygen atmosphere that we depend on was generated by numerous cyanobacteria during the Archaean and Proterozoic Eras.

-Cyanobacteria are aquatic and photosynthetic, meaning they live in the water, and can manufacture their own food. During the process of photosynthesis, energy from sunlight is used together with carbon (iv) oxide and water to form organic molecules that serve as nutrients for the organisms and also oxygen gas is released to the atmosphere.

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Covalent bonds in a molecule absorb radiation in the IR region and vibrate at characteristic frequencies.

vazorg [7]

Answer:

$Frequency=1.6\times 10^{12}\ Hz$

Explanation:

The relation between frequency and wavelength is shown below as:

$c=frequency\times Wavelength
$

c is the speed of light having value $3\times 10^8\ m/s$

Given, Wavelength = 188 μm

Also, 1 μm = $10^{-6}$ nm

So,

Wavelength = $188\times 10^{-6}$ m

Thus, Frequency is:

$Frequency=\frac{c}{Wavelength}$

$Frequency=\frac{3\times 10^8}{188\times 10^{-6}}\ Hz$

$Frequency=1.6\times 10^{12}\ Hz$

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32L —> 32000g —> 727.116 Moles (rounded)

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3 years ago

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a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re

Blababa [14]

Answer: 41.5 mL

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

Given : 59.4 g of $H_2SO_4$ in 100 g of solution

moles of $H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61$

Volume of solution = $\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml$

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.61\times 1000}{54.6ml}=11.2M$

To calculate the volume of acid, we use the equation given by neutralisation reaction:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of stock acid which is $H_2SO_4$

$M_2\text{ and }V_2$ are the molarity and volume of dilute acid which is $H_2SO_4$

We are given:

$M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL$

Putting values in above equation, we get:

$11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL$

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

4 0

3 years ago