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natta225 [31]
2 years ago
8

Do you think that solids can undergo convection? explain

Chemistry
1 answer:
3241004551 [841]2 years ago
3 0
No. Solids by definition undergo conduction and only fluids can undergo convection.
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The diagram shows the scales used for recording temperatures. The labels for the scales are missing. 3 thermometers are oriented
QveST [7]

Answer d

Explanation:

8 0
2 years ago
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A double-replacement reaction takes place when aqueous k2so4 reacts with aqueous pb(no3)2. you would expect the products of this
SpyIntel [72]
A double replacement is when the reactants switch with their opposite charge. For example, AB + CD = AD + BC.

In this case the given is K2SO4 + Pb(NO3)2 = ____ + ____

The answers is PbSO4 and KNO3 would expect the products of this reaction.
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2 years ago
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20 PTS!!! What is the volume of a gas at 90.0°C, if it occupies 1.41 L at 170°C?
Margarita [4]

Answer:

1.72 L

Explanation:

Initial Temperature of the gas T =90°+273= 363 K

Initial Volume of the gas V1= 1.41 L

Final temperature of the gas T2= 170°c +273= 443 K

Final volume of the gas V2= ????

Using Charles law;

V1/T1 = V2/T2

V1T2= V2T1

V2= V1T2/T1

V2= 1.41×443/363

V2= 1.72 L

3 0
2 years ago
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500ml of a buffer solution contains 0.050 mol nahso3 and 0.031
nydimaria [60]

Answer:

The answers are explained below

Explanation:

a)

Given: concentration of salt/base = 0.031

concentration of acid = 0.050

we have

PH = PK a + log[salt]/[acid] = 1.8 + log(0.031/0.050) = 1.59

b)

we have HSO₃⁻ + OH⁻ ------> SO₃²⁻ + H₂O

Moles i............0.05...................0.01.................0.031.....................0

Moles r...........-0.01.................-0.01................0.01........................0.01

moles f...........0.04....................0....................0.041.....................0.01

c)

we will use the first equation but substituting concentration of base as 0.031 + 10ml = 0.031 + 0.010 = 0.041

Hence, we have

PH = PK a + log[salt]/[acid] = 1.8 + log(0.041/0.050) = 1.71

d)

pOH = -log (0.01/0.510) = 1.71

pH = 14 - 1.71 = 12.29

e)

Because the buffer solution (NaHSO3-Na2SO3) can regulate pH changes. when a buffer is added to water, the first change that occurs is that the water pH becomes constant. Thus, acids or bases (alkali = bases) Additional may not have any effect on the water, as this always will stabilize immediately.

4 0
2 years ago
Consider two solutions, the first being 50.0 mL of 1.00 M CuSO4 and the second 50.0 mL of 2.00 M KOH. When the two solutions are
kolbaska11 [484]

Explanation:

Molarity of copper sulfate solution = 1.00 M

Volume of the copper sulfate solution  = 50.0 mL = 0.050 L

Moles of copper sulfate = n

1.00M=\frac{n}{0.050 L}

n = 0.050 L × 1.00 M= 0.050 mol

1 mol of copper sulfate has 1 mol of copper . Then 0.050 mol of copper sulfate has :

1\times 0.050 mol=0.050 mol of copper

a) Mass of 0.050 moles of copper = 0.050 mol × 63.5 g/mol =3.175 g

b) The identity of the compound which formed after the reaction is copper hydroxide.

c) The complete equation for the reaction that occurs when copper sulfate and potassium hydroxide are mixed:

CuSO_4(aq)+2KOH (aq)\rightarrow Cu(OH)_2(s)+K_2SO_4 (aq)

d) CuSO_4(aq)\rightarrow Cu^{2+}(aq) +SO_{4}^{2-}(aq)..[1]

KOH (aq)\rightarrow 2K^+(aq) +OH^-(aq)..[2]

Cu^{2+}(aq) +SO_{4}^{2-}(aq)+2K^+(aq) +2OH^-(aq)\rightarrow Cu(OH)_2(s)+SO_{4}^{2-}(aq)+2K^+(aq)

Common ion both sides are removed. The net ionic equation is given as:

Cu^{2+}(aq) +2OH^-(aq)\rightarrow Cu(OH)_2(s)(aq)

e) Volume of solution after mixing of both solution,V= 50 mL + 50ml = 100 mL

Mass of final solution ,m= 1 mL

Density of solution ,d= 1 g/mL (same as pure water)

m=d\time V=1 g/ml\times 100 mL = 100 g

Heat capacity of the solution = c = 4.186 J/g°C (same as pure water)

Change in temperature of the solution,ΔT = 27.7 °C- 21.5 °C=6.2°C

Q=mc\Delta T

Q=100 g\times 4.186 J/g ^oC\times 6.2^oC=2595.32 J=2.595 kJ

Enthalpy of the reaction = ΔH = \frac{Q}{\text{Moles of copper}}

ΔH = \frac{2.595 kJ}{0.050 mol}=51.90 kJ/mol

The ΔH for the reaction that occurs on mixing is 51.90 kJ/mol.

7 0
2 years ago
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