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2 years ago

put the ions in order from the ion with the fewest oxygen atoms to the ion with the most oxygen atoms.

Chemistry

1 answer:

dezoksy [38]2 years ago

7 0

Answer:

Chloride , hypochlorite, chlorite , chlorate, perchlorate

Chloride ion Cl-

Perchlorate ClO4-

Chlorate ClO3-

Chlorite ClO2-

Hypochlorite ClO-

Explanation:

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Atomic number for bromine

Yuliya22 [10]

Atomic number for bromine is 35

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3 years ago

Which atom has a mass of 12 u

Andrews [41]

Answer:

Carbon-12

Explanation:

4 0

3 years ago

Read 2 more answers

Be sure to answer all parts. calculate δg ocell for the reaction between cr(s) and cu2+(aq). e ocell = 1.08 j/c. enter your answ

Mademuasel [1]

Answer:

$\boxed{-6.29 \times10^{5}\text{ J}}$

Explanation:

Step 1. Determine the cell potential

E°/V

2×[Cr ⟶ Cr³⁺ + 3e⁻] 0.744 V

3×[Cu²⁺ + 2e⁻ ⟶ Cu] 0.3419 V

2Cr + 3Cu²⁺ ⟶ 3Cu + 2Cr³⁺ 1.086 V

Step 2. Calculate ΔG°

$\Delta G^{\circ} = -nFE_{\text{cell}}^{^{\circ}} = -6 \times 96 485 \times 1.086 = \text{-629 000 J}\\\\= \boxed{-6.29 \times10^{5}\text{ J}}$

6 0

3 years ago

What is the product of a force and the time during which the force acts

ELEN [110]

Impulse is the product of a force and the time during which that force acts on a body.

7 0

3 years ago

Calculate the enthalpy of the reaction

harkovskaia [24]

Answer : The enthalpy of the reaction is, -2552 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given enthalpy of reaction is,

$4B(s)+3O_2(g)\rightarrow 2B_2O_3(s)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $B_2O_3(s)+3H_2O(g)\rightarrow 3O_2(g)+B_2H_6(g)$ $\Delta H_A=+2035kJ$

(2) $2B(s)+3H_2(g)\rightarrow B_2H_6(g)$ $\Delta H_B=+36kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_C=-285kJ$

(4) $H_2O(l)\rightarrow H_2O(g)$ $\Delta H_D=+44kJ$

Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :

(when we are reversing the reaction then the sign of the enthalpy change will be change.)

The expression for enthalpy of the reaction will be,

$\Delta H=-2\times \Delta H_A+2\times \Delta H_B-6\times \Delta H_C-6\times \Delta H_D$

$\Delta H=-2(+2035kJ)+2(+36kJ)-6(-285kJ)-6(+44)$

$\Delta H=-2552kJ$

Therefore, the enthalpy of the reaction is, -2552 kJ/mole

4 0

3 years ago