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3 years ago

Which substances are released during cellular respiration?

1 answer:

8 0

The formula for cellular respiration is the opposite of photosynthesis in a way of remembering it. Glucose (C6H12O6) + oxygen to break it down makes 2 waste products. 6H2O (water) and 6CO2 (carbon dioxide) The most important product would be ATP that the cells use for energy, but I am assuming you wanted to know about what is released. Example: When you exercise, you breath in more oxygen which breaks down more glucose which makes more ATP. That’s why you feel tired. And you start to breath out more and release more CO2 as a waste product being released. Hope this helped!

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What are two ways in which white blood cells fight pathgons that have entered the body

crimeas [40]

Answer:

Answer c i believe

Explanation:

3 0

3 years ago

When 3.00 g of sulfur are combined with 3.00 g of oxygen, 6.00 g of sulfur dioxide (SO2) are formed. What mass of oxygen would b

bazaltina [42]

Actually, we can answer the problem even without the first statement. All we have to do is write the reaction for the production of sulfur trioxide.

2 S + 3 O₂ → 2 SO₃

The stoichiometric calculations is as follows:

6 g S * 1 mol/32.06 g S = 0.187 mol S
Moles O₂ needed = 0.187 mol S * 3 mol O₂/2 mol S = 0.2805 mol O₂
Since the molar mas of O₂ is 32 g/mol,
Mass of O₂ needed = 0.2805 mol O₂ * 32 g/mol = 8.976 g O₂

4 0

3 years ago

How many pints of a 30% sugar solution must be added to a 5 pint of a 5% sugar solution to obtain a 20% sugar solution?

ser-zykov [4K]

You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.

You can use a modified dilution formula to calculate the volume of 30 % sugar.

V_1×C_1 + V_2×C_2 = V_3×C_3

Let the volume of 30 % sugar = x pt. Then the volume of the final 20 % sugar = (5 + x ) pt

(x pt×30 % sugar) + (5 pt×5 % sugar) = (x + 5) pt × 20 % sugar

30x + 25 = 20x + 100

10x = 75

x = 75/10 = 7.5

5 0

3 years ago

A certain electromagnetic wave a wavelength of 415 nm. What is the frequency of the wave in hz?

Kaylis [27]

Answer:

The frequency of the electromagnetic wave is 7.22891566 × 10¹⁴ Hz

Explanation:

The wavelength of the electromagnetic wave, λ = 415 nm

The speed of an electromagnetic wave, c ≈ 3.0 × 10⁸ m/s

Given that an electromagnetic wave is a periodic wave, we have;

The speed of the electromagnetic wave, c = f×λ

Where;

f = The frequency of the electromagnetic wave

Therefore, we have;

f = c/λ

From which we have;

f = (3.0 × 10⁸ m/s)/(415 nm) = 7.22891566 × 10¹⁴ /s = 7.22891566 × 10¹⁴ Hz

The frequency of the electromagnetic wave, f = 7.22891566 × 10¹⁴ Hz

8 0

3 years ago

Titanium has an HCP crystal structure, a c/a ratio of 1.669, an atomic weight of 47.87 g/mol, and a density of 4.51 g/cm3. Compu

IrinaK [193]

Answer : The atomic radius for Ti is, $1.45\times 10^{-8}cm$

Explanation :

Atomic weight = 47.87 g/mole

Avogadro's number $(N_{A})=6.022\times 10^{23} mol^{-1}$

First we have to calculate the volume of HCP crystal structure.

Formula used :

$\rho=\frac{Z\times M}{N_{A}\times V}$ .............(1)

where,

$\rho$ = density = $4.51g/cm^3$

Z = number of atom in unit cell (for HCP = 6)

M = atomic mass = 47.87 g/mole

$(N_{A})$ = Avogadro's number

V = volume of HCP crystal structure = ?

Now put all the values in above formula (1), we get

$4.51g/cm^3=\frac{6\times (47.87g/mol)}{(6.022\times 10^{23}mol^{-1}) \times V}$

$V=1.06\times 10^{-22}cm^3$

Now we have to calculate the atomic radius for Ti.

Formula used :

$V=6R^2c\sqrt{3}$

Given:

c/a ratio = 1.669 that means, c = 1.669 a

Now put (c = 1.669 a) and (a = 2R) in this formula, we get:

$V=6R^2\times (1.669a)\sqrt{3}$

$V=6R^2\times (1.669\times 2R)\sqrt{3}$

$V=(1.669)\times (12\sqrt{3})R^3$

Now put all the given values in this formula, we get:

$1.06\times 10^{-22}cm^3=(1.669)\times (12\sqrt{3})R^3$

$R=1.45\times 10^{-8}cm$

Therefore, the atomic radius for Ti is, $1.45\times 10^{-8}cm$

3 0

3 years ago