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Chemistry-

General

Easy

Question

Which will give $N subscript 2 end subscript O$ on heating?

$N H_{4} N O_{2}$
$N H_{4} N O_{3}$
$N a N O_{3}$
$L i N O_{3}$

The correct answer is: $N H subscript 4 end subscript N O subscript 3 end subscript$

$N H subscript 4 end subscript N O subscript 3 end subscript rightwards arrow N subscript 2 end subscript O plus 2 H subscript 2 end subscript O$

Related Questions to study

Some of alkali metal salts are coloured. $N a subscript 2 end subscript C r O subscript 4 end subscript minus$ yellow, $K M n O subscript 4 end subscript minus$ pink, $K subscript 2 end subscript M n O subscript 4 end subscript minus$ green. It is due to

Some of alkali metal salts are coloured. $N a subscript 2 end subscript C r O subscript 4 end subscript minus$ yellow, $K M n O subscript 4 end subscript minus$ pink, $K subscript 2 end subscript M n O subscript 4 end subscript minus$ green. It is due to

Chemistry-General

The equation of the hyperbola whose foci are (6, 4) and (–4, 4) and eccentricity 2 is given by

The equation of the hyperbola whose foci are (6, 4) and (–4, 4) and eccentricity 2 is given by

Mg and Li are similar in their properties due to

Mg and Li are similar in their properties due to

Chemistry-General

parallel

When an electron makes a transition from $left parenthesis n plus 1 right parenthesis$ state to $n$ state, the frequency of emitted radiations is related to $n$ according to $left parenthesis n much greater-than 1 right parenthesis$

When an electron makes a transition from $left parenthesis n plus 1 right parenthesis$ state to $n$ state, the frequency of emitted radiations is related to $n$ according to $left parenthesis n much greater-than 1 right parenthesis$

Chemistry-General

The binding energy of an element is 64 Mev. If BE/nucleon is 6.4, then the number of nucleons are

The binding energy of an element is 64 Mev. If BE/nucleon is 6.4, then the number of nucleons are

Chemistry-General

The ionization potential hydrogen atom is $13.6 blank e V$ . The energy required to remove an electron in the $n equals 2$ state of the hydrogen atom is

The ionization potential hydrogen atom is $13.6 blank e V$ . The energy required to remove an electron in the $n equals 2$ state of the hydrogen atom is

Chemistry-General

parallel

In nuclear reaction, $blank subscript 3 end subscript superscript 7 end superscript L i plus blank subscript 1 end subscript superscript 1 end superscript H rightwards arrow 2 blank subscript 2 end subscript superscript 4 end superscript H e$ The mass loss is nearly 0.02 amu. Hence, the energy released (in units of million kcal/mol) in the process is approximately

In nuclear reaction, $blank subscript 3 end subscript superscript 7 end superscript L i plus blank subscript 1 end subscript superscript 1 end superscript H rightwards arrow 2 blank subscript 2 end subscript superscript 4 end superscript H e$ The mass loss is nearly 0.02 amu. Hence, the energy released (in units of million kcal/mol) in the process is approximately

Chemistry-General

There are $0.618 blank mu g blank$ of $blank to the power of 206 end exponent P b$ and $0.238 blank mu g blank$ of $blank to the power of 238 end exponent U$ in a rock. If $T subscript 50 end subscript$ of $blank to the power of 238 end exponent U$ is $1.5 cross times 10 to the power of 9 end exponent y r comma$ age of the rock is

There are $0.618 blank mu g blank$ of $blank to the power of 206 end exponent P b$ and $0.238 blank mu g blank$ of $blank to the power of 238 end exponent U$ in a rock. If $T subscript 50 end subscript$ of $blank to the power of 238 end exponent U$ is $1.5 cross times 10 to the power of 9 end exponent y r comma$ age of the rock is

Chemistry-General

The latus-rectum of the hyperbola $16 x to the power of 2 end exponent minus 9 y to the power of 2 end exponent equals$ $144$ , is

The latus-rectum of the hyperbola $16 x to the power of 2 end exponent minus 9 y to the power of 2 end exponent equals$ $144$ , is

parallel

Let $stack a with rightwards arrow on top equals 2 stack i with hat on top plus stack j with hat on top minus 2 stack k with hat on top$ and $stack b with rightwards arrow on top equals stack i with hat on top plus stack j with hat on top$ . If $stack c with rightwards arrow on top$ is a vector such that $stack a with rightwards arrow on top. stack c with rightwards arrow on top equals open vertical bar stack c with rightwards arrow on top close vertical bar comma open vertical bar stack c with rightwards arrow on top minus stack a with rightwards arrow on top close vertical bar$ = $2 square root of 2$ and the angle between $open parentheses stack a with rightwards arrow on top cross times stack b with rightwards arrow on top close parentheses$ and $stack c with rightwards arrow on top$ is 30⁰, then find the value of $open vertical bar open parentheses stack a with rightwards arrow on top cross times stack b with rightwards arrow on top close parentheses cross times stack c with rightwards arrow on top close vertical bar$ .

Let $stack a with rightwards arrow on top equals 2 stack i with hat on top plus stack j with hat on top minus 2 stack k with hat on top$ and $stack b with rightwards arrow on top equals stack i with hat on top plus stack j with hat on top$ . If $stack c with rightwards arrow on top$ is a vector such that $stack a with rightwards arrow on top. stack c with rightwards arrow on top equals open vertical bar stack c with rightwards arrow on top close vertical bar comma open vertical bar stack c with rightwards arrow on top minus stack a with rightwards arrow on top close vertical bar$ = $2 square root of 2$ and the angle between $open parentheses stack a with rightwards arrow on top cross times stack b with rightwards arrow on top close parentheses$ and $stack c with rightwards arrow on top$ is 30⁰, then find the value of $open vertical bar open parentheses stack a with rightwards arrow on top cross times stack b with rightwards arrow on top close parentheses cross times stack c with rightwards arrow on top close vertical bar$ .

The equation of the line passing through the point (1, -2, 3) and parallel to the line x – y + 2z = 5 and 3x + y + z = 6 is

The equation of the line passing through the point (1, -2, 3) and parallel to the line x – y + 2z = 5 and 3x + y + z = 6 is

If (1 + x + x²)ⁿ = a₀ + a₁x + a₂x² + ………+a_2nx²ⁿ, then the value of the determinant $open vertical bar table row cell a subscript n minus 3 end subscript end cell cell a subscript n minus 1 end subscript end cell cell a subscript n plus 1 end subscript end cell row cell a subscript n minus 6 end subscript end cell cell a subscript n minus 3 end subscript end cell cell a subscript n plus 3 end subscript end cell row cell a subscript n minus 14 end subscript end cell cell a subscript n minus 7 end subscript end cell cell a subscript n plus 7 end subscript end cell end table close vertical bar$ is always

If (1 + x + x²)ⁿ = a₀ + a₁x + a₂x² + ………+a_2nx²ⁿ, then the value of the determinant $open vertical bar table row cell a subscript n minus 3 end subscript end cell cell a subscript n minus 1 end subscript end cell cell a subscript n plus 1 end subscript end cell row cell a subscript n minus 6 end subscript end cell cell a subscript n minus 3 end subscript end cell cell a subscript n plus 3 end subscript end cell row cell a subscript n minus 14 end subscript end cell cell a subscript n minus 7 end subscript end cell cell a subscript n plus 7 end subscript end cell end table close vertical bar$ is always

parallel

India play two matches each with West Indies and Australia .In any match the probabilities of India getting points 0, 1 and 2 are 0.45, 0.05 and 0.50 respectively. Assuming that the outcomes are independent, the probability of India getting at least 7 points is

India play two matches each with West Indies and Australia .In any match the probabilities of India getting points 0, 1 and 2 are 0.45, 0.05 and 0.50 respectively. Assuming that the outcomes are independent, the probability of India getting at least 7 points is

If the eccentricities of the hyperbolas $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction minus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1$ and $fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction minus fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction equals 1$ be e and $e subscript 1 end subscript$ , then $fraction numerator 1 over denominator e to the power of 2 end exponent end fraction plus fraction numerator 1 over denominator e subscript 1 end subscript superscript 2 end superscript end fraction equals$

If the eccentricities of the hyperbolas $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction minus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1$ and $fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction minus fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction equals 1$ be e and $e subscript 1 end subscript$ , then $fraction numerator 1 over denominator e to the power of 2 end exponent end fraction plus fraction numerator 1 over denominator e subscript 1 end subscript superscript 2 end superscript end fraction equals$

The equation of the normal at the point (2, 3) on the ellipse $9 x to the power of 2 end exponent plus 16 y to the power of 2 end exponent equals 180$ , is

The equation of the normal at the point (2, 3) on the ellipse $9 x to the power of 2 end exponent plus 16 y to the power of 2 end exponent equals 180$ , is

parallel

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