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Maksim231197 [3]
3 years ago
8

How can ocean acidification affect us humans

Chemistry
1 answer:
Elan Coil [88]3 years ago
3 0
These changes are impairing the ocean's capacity to provide food, protect livelihoods, maintain clean water, and recover from environmental stresses like severe storms.
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2NaHCO3-->Na2CO3+H2O+CO2
algol [13]
I’m not sure what kind of answer you’re looking for, but if you’re supposed to translate it into a sentence it would be two moles of sodium bicarbonate(baking soda) decompose unto one mole of sodium carbonate(soda ash) plus one mole of water and one mole of carbon monoxide gas
8 0
3 years ago
Where is most of the mass of an atom? •
yaroslaw [1]
B. inside the nucleus
6 0
3 years ago
F a sample of butene (C4H8) that has a mass of 136.6 g is combusted in excess oxygen, what is the mass of CO2 that is produced?
alukav5142 [94]

The balanced chemical reaction will be:

C4H8 + 6 O2 --> 4 CO2 + 4 H2O

We are given the amount of butene being combusted. This will be our starting point.

136.6 g C4H8 (1 mol  C4H8/ 56.11 g C4H8) (4 mol CO2/1 mol <span>C4H8</span>) ( 44.01 g CO2/ 1 mol CO2) = 428.6 g CO2
7 0
3 years ago
Read 2 more answers
Which of the following is not true about elements? 1. Elements are made of atoms of the same type. 2. Each element has a unique
Marat540 [252]

Sorry If This Is Late***

<u>Elements can't be broken down</u>, into a simplre set of properties. They are one strict unit & cannot be broken down, however they can be added together to make a compound.

HOPE THIS HELPS & good luck <3 !!!!!

7 0
2 years ago
Two hydrogen atoms collide in a head on collision and end up with zero kinetic energy. Each then emits a photon of 121.6 nm (n=2
Zinaida [17]

Explanation:

Expression for the kinetic energy is as follows.

         K.E = \frac{1}{2}mv^{2}

Now, total kinetic energy will be as follows.

    K.E = 2 \times \frac{1}{2}mv^{2} = m \times v^{2}

Since, this energy converts into electromagnetic radiation of wavelength 121.6 nm.

Relation between energy and photon is as follows.

   Energy of photon = \frac{hc}{\lambda}

                                = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{121.6 \times 10^{-9}}

                                 = 1.63 \times 10^{-18} J

    m \times v^{2} = 1.63 \times 10^{-18}

          v = \sqrt{\frac{1.63 \times 10^{-18}}{1.67 \times 10^{-27}}

             = 3.12 \times 10^{4} m/s

Thus, we can conclude that atoms were moving at a speed of 3.12 \times 10^{4} m/s before the collision.

8 0
3 years ago
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