A NOR gate outputs true when both inputs are false.

A NOR gate outputs 1 only when both inputs are 0. Therefore, if both inputs are low (0), the output is 1.

A NOT gate inverts the input. Therefore, NOT 0 = 1.

A NOT gate outputs false when the input is true.

An AND gate outputs 1 only when both inputs are 1. Since one input is 0, the output is 0.

An AND gate outputs 1 only when both inputs are high. Since both A and B are 0, the output is 0.

An AND gate outputs 1 only when both inputs are high. Since A = 0, the output is 0.

An OR gate outputs true if at least one input is true.

An OR gate outputs 0 only when both inputs are low. Therefore, the output is 0.

An XOR gate outputs false when both inputs are the same.

A AND B = 0 AND 1 = 0. Therefore, NOT (0) = 1.

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In NaClO3, chlorine has an oxidation state of +5.

In NaClO3, the oxidation state of chlorine is +5. Sodium is +1 and each oxygen is -2, leading to the equation: +1 + x + 3(-2) = 0, solving gives x = +5.

In K2Cr2O7, the total oxidation state of the two potassium ions and seven oxygen ions is -12, leading to +6 for each chromium.

In K2Cr2O7, the total charge from potassium (K) is +2, and oxygen (O) contributes -14. Let the oxidation state of Cr be x. The equation is 2x - 14 + 2 = 0, solving gives x = +6.

In K2[Cr2O7], the overall charge is 0. Each potassium (K) has a +1 charge, and each oxygen (O) has a -2 charge. Solving for chromium gives an oxidation state of +6.

Let the oxidation state of Cr be x. The equation is x + 4(0) + 2(-1) = +1 (from Cl). Solving gives x = +3.

Let the oxidation state of Cr be x. The equation is x + 4(0) + 2(-1) = +1. Solving gives x = +3.

Let the oxidation state of Cr be x. The equation is x + 4(0) + 2(-1) = +1. Solving gives x = +3.

The complex is neutral, and the three chloride ions contribute -3 charge. Therefore, the oxidation state of Cr must be +3.

In water (H2O), the oxidation state of hydrogen is +1.

In metal hydrides, hydrogen has an oxidation state of -1.

In most of its compounds, hydrogen has an oxidation state of +1.

In KMnO4, the oxidation state of manganese is +7.

In NH4+, nitrogen has an oxidation state of -3.

In NH4+, nitrogen has an oxidation state of -3. Each hydrogen is +1, leading to the equation: x + 4(+1) = +1, solving gives x = -3.

In NH4Cl, the oxidation state of nitrogen is -3, as it is bonded to four hydrogens.

In NO3^-, nitrogen has an oxidation state of +5.

In H3PO4, phosphorus has an oxidation state of +5.