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katrin [286]
3 years ago
9

Plz answer for brainliest trolls will get deleted

Chemistry
1 answer:
Roman55 [17]3 years ago
3 0

Answer:

A; C6H12O6 + 6O2

Explanation:

You can easily differentiate between the reactants and the products, as the reactants are on the left, while the products are on the right.

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The ksp of calcium carbonate, caco3, is 3.36 × 10-9 m2. calculate the solubility of this compound in g/l.
maw [93]
CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
                       CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial                Y                   -                 -
Change           -X                  +X              +X
Equilibrium      Y-X                 X                X

Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²

                Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
                    X = 5.79 x 10⁻⁵ M

Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
                                                     = 5.79 x 10⁻⁵ mol/L

Molar mass of CaCO₃ = 100 g mol⁻¹

Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
                                                 = 5.79 x 10⁻³ g/L

7 0
3 years ago
A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el
musickatia [10]

Answer:

T_2=335.42K=62.27^oC

Explanation:

Hello,

In this case, by using the general gas law, that allows us to understand the pressure-volume-temperature relationship as shown below:

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

Thus, solving for the temperature at the end (considering absolute units of Kelvin), we obtain:

T_2=\frac{P_2V_2T_1}{P_1V_1}=\frac{1.8L*0.75atm*(25+273.15)K}{1.2L*1.0atm} \\\\T_2=335.42K=62.27^oC

Best regards.

4 0
3 years ago
Read 2 more answers
I will mark brainly , please help !
topjm [15]
...a metal atom will *lose* electrons to form a *positive* cation and a nonmetal atom will *accept* electrons to form an *negative* anion.
7 0
3 years ago
G.com what is the mass of a gold bar that is 7.379 × 10–4 m3 in volume
tiny-mole [99]
<span>7.379 * 10^(-4) is measured, hence prone to error, either human error or via measuring device. In this case,
100 cm = 1 m is written in stone and is unquestionable.
 The density of the gold is 19.3 g/cm^3 and could be an approximation.
 The approximation is good to at least one night.</span>
5 0
2 years ago
Choose the ingredients needed for nuclear fusion. Check all that apply. energy helium gas high temperatures hydrogen gas low pre
Igoryamba

answer:

high temperatures

hydrogen gas

7 0
3 years ago
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