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Ne4ueva [31]
3 years ago
11

Whats photosynthesis

Chemistry
1 answer:
s344n2d4d5 [400]3 years ago
4 0
It is a process of a plant taking in sun rays and changing it into glucose ur sugar  
The formula for this process is :
6CO_2 + 6H_2O -----> C6H12O6+6O_2
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Astatine-210 has a half-life of 8.08 days. What fraction of a sample of astatine-210 is left unchanged after 16.16 days?
Flauer [41]

Answer:

0

Explanation:

Given parameters:

Half-life  = 8.08days

Unknown:

What fraction is left unchanged after 16.16days = ?

Solution:

The half - life of a substance is the time taken for the half of a radioactive material to decay to half.

 

  Day 0           Day 8.08         Day 16.16

   100%                 50%                 0%        Parent

    0%                     50%                100%     Daughter

After 16.16 days, non of the original sample will remain unchanged.

6 0
3 years ago
Write a caption for a photo of a polluted city.​
arlik [135]

Answer:

ok

Explanation:

this is a picture of a city that we have made worse everyday

5 0
3 years ago
2C3H7OH + 9O2 -> 6CO2 + 8H2O
MAVERICK [17]

Answer:

Up to 18\; \rm mol \rm CO_2.

Explanation:

The equation for this reaction: 2\; \rm C_3 H_7 OH + 9\; O_2 \to 6\; CO_2 + 8\; H_2O is indeed balanced.

Notice that the coefficient of \rm C_3 H_7 OH is 2 while that of \rm CO_2 is 6.

Therefore, each unit of this reaction would consume 2 units of \rm C_3H_7OH while producing 6 units of \rm CO_2. In other words, this reaction would produce three times as much \rm CO_2\! as it consumes \rm C_3H_7OH\!.

If all that 6\; \rm mol of \rm C_3H_7OH molecules are actually consumed through this reaction, three times as many \rm CO_2 molecules would be produced. That corresponds to 3 \times 6\; \rm mol = 18\; \rm mol of \rm CO_2\! molecules.

6 0
3 years ago
Why is HBrO a weak acid
Juli2301 [7.4K]
Because it contains the weak base BrO which makes it a weak acid even though HBr is a strong acid
6 0
3 years ago
For the reaction IO3–(aq) + 5 I–(aq) + 6 H+(aq) → 3 I2(aq) + 3 H2O(l) the rate of disappearance of I–(aq) at a particular time a
scoray [572]

Answer: The rate of appearance of I_2(aq) is 1.44\times 10^{-3}mol/Ls

Explanation:

Rate of a reaction is defined as the rate of change of concentration per unit time.

Thus for reaction:

IO_3^-(aq)+5I^-(aq)+6H^+(aq)\rightarrow 3I_2(aq)+3H_2O(l)

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate=-\frac{1d[I^-]}{5dt}=+\frac{d[I_2]}{3dt}

Given: -\frac{d[I^-]}{dt}] = 2.4\times 10^{-3}mol/Ls

+\frac{d[I_2]}{dt}=-\frac{3d[I^-]}{5dt}=-\frac{3}{5}\times 2.4\times 10^{-3}mol/Ls=1.44\times 10^{-3}mol/Ls

The rate of appearance of I_2(aq) is 1.44\times 10^{-3}mol/Ls

6 0
3 years ago
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