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3 years ago

As matter changes, what happens to the energy?

Chemistry

1 answer:

OlgaM077 [116]3 years ago

4 0

Energy, Temperature, and Changes of State

Matter either loses or absorbs energy when it changes from one state to another. For example, when matter changes from a liquid to a solid, it loses energy. The opposite happens when matter changes from a solid to a liquid.

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One way to heat a gas is to compress it. a gas at 1.00 atm at 25.0°c is compressed to one tenth of its original volume, and it r

Tanzania [10]

(p1)(V1)/(T1) = (p2)(V2)/(T2)
(1.00 atm)(V) / (273 + 25K) = (40.0 atm)(V/10) / (273 + T)
273 + T = (40.0)(1/10)(273 + 25K) / (1.00)
T = 919°C

5 0

3 years ago

Chloroform, CHCl3, reacts with chlorine, Cl2, to form carbon tetrachloride, CCl4, and hydrogen chloride, HCl. In an experiment 2

erastovalidia [21]

Answer:

Chloroform= limiting reactant

0.209mol of CCl4 is formed

And 32.186g of CCl4 is formed

Explanation:

The equation of reaction

CHCl3 + Cl2= CCl4 + HCl

From the equation 1 mol of

CHCl3 reacts with 1mol Cl2 to yield 1mol of CCl4

From the question

25g of CHCl3 really with Cl2

Molar mass of CHCl3= 119.5

Molar mass of Cl2 = 71

Hence moles of CHCl3= 25/119.5 = 0.209mol

Moles of Cl2 = 25/71 = 0.352mol

Hence CHCl3 is the limiting reactant

Since 1 mole of CHCl3 gave 1mol of CCl4

It implies that 0.209moles of CHCl3 will also give 0.209mol of CCl4

Mass of CCl4 formed = moles× molar mass= 0.209×154= 32.186g

6 0

3 years ago

Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta

BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, $C_{Ag}$ = 78 wt%

concentration of Pd, $C_P_d$ = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, $A_A_g$ = 107.87 g/mol

atomic weight of iron, $A_P_d$ = 106.4 g/mol

Step 1: determine the average density of the alloy

$\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }$

$\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3$

Step 2: determine the average atomic weight of the alloy

$A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }$

$A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole$

Step 3: determine unit cell volume

$V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}$

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

$V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell$

Step 4: determine the unit cell edge length

Vc = a³

$a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm$ = 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0

3 years ago

What do electrons and neutrons have in common?

bearhunter [10]

█ Answer █

Electrons and neutrons are both sub-atomic particles.

Hope that helps! ★ If you have further questions about this question or need more help, feel free to comment below or leave me a PM. -UnicornFudge aka Nadia

8 0

3 years ago

Read 2 more answers

A atom neutral (one with 0 extra charge)?

telo118 [61]

Nuetrons=neutral
protons =positive
electrons=negative

8 0

3 years ago