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3 years ago

What produces and affects the strength of a magnetic field

Chemistry

1 answer:

AnnyKZ [126]3 years ago

6 0

Answer:

Factors Affecting the Strength of the Magnetic Field of an Electromagnet: Factors that affect the strength of electromagnets are the nature of the core material, strength of the current passing through the core, the number of turns of wire on the core and the shape and size of the core.

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Which species is the conjugate base of an ammonium ion, NH4+?

SIZIF [17.4K]

Answer:

The conjugated base of ammonium ion $NH_{4}^{+}$ is $NH_{5}^{2+}$

Explanation:

A conjugate base is obtained by addition of proton $H^{+}$ . In order to get the conjugate base of $NH_{4}^{+}$ , we need to add a proton to ammonium ion. We will get $NH_{5}^{2+}$ after addition of proton to ammonium ion.

Hence, the conjugate base of $NH_{4}^{+}$ is $NH_{5}^{2+}$

3 0

4 years ago

When both –NH2 and –COOH replace hydrogen atoms on the same carbon atom, a(n) ____ is formed. a. amine c. carboxylic acid b. ami

Mice21 [21]

$-NH_{2}$ represents amine functional group and $COOH$ represents carboxylic acid functional group. When these both functional groups are substituted onto a single carbon atom, then amino acid is formed.

The organic compound consists of $-NH_{2}$ functional group and $COOH$ functional group with a side chain is said to be an amino acid.

A: amine group is $-NH_{2}$ .

C: carboxylic acid group is $COOH$

B: amino acid represents $-NH_{2}$ functional group and $COOH$ functional group which are substituted onto a single carbon atom.

D: carbohydrate: the general formula of carbohydrate is $(CH_{2} O)_{x}$ .

Hence, amino acid is the correct answer.

4 0

3 years ago

Some SO2 and O2 are mixed together in a flask at 1100 K in such a way ,that at the instant of mixing, their partial pressures ar

kakasveta [241]

Answer:

The answer is " $\bold{0.525\ \ atm^{-1}}$ "

Explanation:

Given equation:

$2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g)$

Given value:

$\Delta rH =-198.2 \ \ \frac{KJ}{mol}$

$Kp=1100 \ K$

$\Delta x = 2-(2+1)\\\\$

$= 2-(2+1)\\\\= 2-(3)\\\\= -1$

$\left\begin{array}{cccc}I\ (atm)&1&0.5&0\\C\ (atm)&2x&-x&2x\\E\ (atm) &1-2x&0.5-x&2x\end{array}\right$

calculating the total pressure on equilibrium= $(1-2x)+(0.5-x)+2x \ atm\\\\$

$= 1-2x+0.5-x+2x\\\\= 1+0.5-x\\\\=1.5-x\\$

$\therefore \\\\\to 1.50-x= 1.35 \\\\\to 1.50-1.35= x\\\\\to 0.15= x\\\\ \to x= 0.15\\\\$

calculating the pressure in $So_2$ :

$= (1-2 \times 0.15)$

$= 1-0.30 \\\\ =0.70 \ atm$

calculating the pressure in $O_2$ :

$= (0.5- 0.15)\\\\= 0.35 \ atm \\$

calculating the pressure in $So_3$ :

$= (2 \times 0.15)\\\\= (.30) \ atm \\\\$

Calculating the Kp at 1100 K:

$= \frac{(Pressure(So_3))^2}{(Pressure(So_2))^2 \times Pressure(O_2)}\\\\= \frac{0.30^2}{0.70^2 \times 0.35}\\\\= \frac{0.30 \times 0.30 }{0.70\times 0.70 \times 0.35}\\\\= \frac{0.09 }{0.49\times 0.35} \\\\= \frac{0.09 }{0.1715} \\\\= 0.5247 \ \ or \ \ 0.525 \ \ atm^{-1} \\\\$