All categories

3 years ago

On a phrase diagram or heating curve, the freezing point is the same as what ?

Chemistry

1 answer:

yuradex [85]3 years ago

8 0

Answer:

Pelting point.

Explanation:

Freezing point is the same is the melting point.

You might be interested in

Please help me vote you brainiest

xxTIMURxx [149]

Answer: D

Explanation: P = W/t

W = F d

When dividing, if you have a large number over a small, you get a large number. We want a small number, so we want the opposite for our P : W/ t formula. We want a small Work and a large time.

Right off the bat, the two with the least time aren't right then.

From there, we can estimate how much work it would take using the W = F d equation. If there is a small distance, there is less work, which is what we want to achieve a small work over large time.

So, the choice with the most time and the least distance should be correct.

3 0

3 years ago

In the copper – silver nitrate lab Copper medals and silver nitrate solution reacted to produce silver metal and copper (II) nit

Nataly [62]

Answer:

Percent Yield = 97.75 %

Explanation:

1 MOLE = It is equal to the molar mass of the substance

1 mole of Cu = 63.54 g (Molar Mass of Cu = 63.54 g/mole)

1 mole of AgNO3 = 170 g (Molar Mass of AgNO3 = 170 g/mol)

Given Mass of AgNO3 = 1.41 g

Given Mass of Cu = 2.93 g

<em><u>Second step : Find the limiting Reagent (which is in less amount) </u></em>

Balanced Chemical equation :

$Cu + 2AgNO_{3} \rightarrow Cu(NO_{3})_{2}+2Ag$

This means

1 mole of Cu will react with = 2 mole of AgNO3

63.54 g of Cu reacts with = 2 x 170 g of AgNO3

1 g of Cu reacts with = (2 x 170)/63.54 of AgNO3

$= \frac{170\times 2}{63.54}$

= 5.35 g of AgNO3

2.93 g should reacts with = 2.93 x 5.35 = 15.67 g of AgNO3

Available AgNO3 = 1.41 g

So , AgNO3 is less than required = limiting reagent

Now the reaction occur 1.41 g of AgNO3

Now, Limiting reagent will decide How much Silver(Ag) Metal will form

2 mole of AgNO3 will produce = 2 mole of Ag

1 mole of AgNO3 will produce = 1 mole of Ag

170 g AgNO3 will produce = 107.86 mole of Ag(Molar mass of Ag = 107.86)

1 g AgNO3 will produce =

$\frac{107.86}{170}$

1.41 g of AgNO3 will produce =

$\frac{107.86\times 1.41}{170}$

= 0.89 g

$Yield\ Percent = \frac{Actual\ Yield}{Theoritical\ Yield}\times 100$

Actual yield = 0.87 g

Theoritical yield = 0.89 g

$Yield\ Percent = \frac{0.87}{0.89}\times 100$

Percent Yield = 97.75 %

5 0

3 years ago

Phosphorous +Dilute hydrochloric acid

butalik [34]

Phosphorus trichloride is the precursor to organophosphorus compounds that contain one or more P(III) atoms, most notably phosphites and phosphonates. ... PCl3 reacts vigorously with water to form phosphorous acid, H3PO3 and HCl: PCl3 + 3 H2O → H3PO3 + 3 HCl.

6 0

3 years ago

What is the break down of food into energy

AnnyKZ [126]

it is nutrients that's it

3 0

4 years ago

Na + H2 → NaCl balanced equation

hodyreva [135]

Answer:

..............

........