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2 years ago

True or False: A nail with a wire wrapped around it is an electromagnet.

2 answers:

7 0

Answer:Search ResultsBy simply wrapping wire that has an electrical current running through it around a nail, you can make an electromagnet. When the electric current moves through a wire, it makes a magnetic field. ... You can make a temporary magnet by stroking apiece of iron or steel (such as a needle) along with a permanent magnet.

Explanation:

Lina20 [59]2 years ago

3 0

Answer:

It is True

Explanation:

By simply wrapping wire that has an electrical current running through it around a nail, you can make an electromagnet. When the electric current moves through a wire, it makes a magnetic field. If you coil the wire around and around, it will make the magnetic force stronger, but it will still be pretty weak.

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The chemical bonding in sodium phosphate, Na3PO4, is classified as:

fiasKO [112]

Ionic bonding between sodium and phosphate ions.

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2 years ago

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Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th

mario62 [17]

Answer: The freezing point of solution is 2.6°C

Explanation:

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$\Delta T_f$ = $\text{Freezing point of pure solution}-\text{Freezing point of solution}$

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point depression constant = 5.12 K/m = 5.12 °C/m

$m_{solute}$ = Given mass of solute (anthracene) = 7.99 g

$M_{solute}$ = Molar mass of solute (anthracene) = 178.23 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC$

Hence, the freezing point of solution is 2.6°C

8 0

2 years ago

How many moles of oxygen are in 52.5 grams of oxygen gas?

adell [148]

An oxygen gas is a diatomic molecule which means that each molecule is composed of 2 atoms. Its symbol is O2.

Each oxygen atom has a molar mass of 16 g/mol. The molar mass of oxygen gas is calculated below,
molar mass = 2 x (16 g/mol) = 32 g/mol

To determine the number of moles in 52.5 grams of oxygen, divide the given mass by the calculated molar mass.

n = 52.5 grams / (32 gram/ mol)
n = 1.64 moles

Thus, there are 1.64 moles of oxygen gas.

3 0

2 years ago

What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as a buffer made from 475 mL of 0.

Hitman42 [59]

Explanation:

The given data is as follows.

[HCOOH] = 0.2 M, [NaOH] = 2.0 M,

V = 500 ml, [Benzoic acid] = 0.2 M

First, we will calculate the number of moles of benzoic acid as follows.

No. of moles of benzoic acid = Molarity × Volume

= $2 \times 0.475$

= 0.095 mol

And, moles of NaOH present in the solution will be as follows.

No. of moles of NaOH = Molarity × Volume

= $2 \times 0.025$

= 0.05 mol

Hence, the ICE table for the chemical equation will be as follows.

$C_{6}H_{5}COOH + NaOH \rightarrow C_{6}H_{5}COONa + H_{2}O$

Initial: 0.095 0.05 0 0

Equlbm: (0.095 - 0.05) 0 0.05

pH = $pK_{a} + log \frac{Base}{Acid}$

= $4.2 + log \frac{0.05}{0.045}$

= 4.245

For,

$HCOOH + NaOH \rightarrow HCOONa + H_{2}O$

Initial: 0.2x 2(0.5 - x) 0

Equlbm: 0.2x - 2(0.5 - x) 0 2(0.5 - x)

As,

pH = $pK_{a} + log \frac{Base}{Acid}$

4.245 = 3.75 + $log \frac{Base}{Acid}$

$log \frac{Base}{Acid}$ = 0.5

$\frac{Base}{Acid}$ = 3.162

Now,

$\frac{2(0.5 - x)}{0.2x - 2(0.5 - x)}$ = 3.162

x = 0.464 L

Volume of NaOH = (0.5 - 0.464) L

= 0.036 L

= 36 ml (as 1 L = 1000 mL)

And, volume of formic acid is 464 mL.

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3 years ago

Draw the structure of the organic product(s) of the grignard reaction between phosgene (clcocl) and excess phenylmagnesium bromi

Jobisdone [24]

Phosgene on reacting with phenylmagnesium bromide generates benzoyl chloride.

Since, phenylmagnesium bromide is added in excess. It would further react with benzoyl chloride to form benzophenone.

Benzophenone on further reacting with phenylmagnesium bromide, and aqueous treatment, gives triphenylmethanol.

Entire reaction pathways is shown below:

8 0

3 years ago

Read 2 more answers