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3 years ago

What will replace aluminum in a single replacement reaction

Chemistry

2 answers:

Liula [17]3 years ago

7 0

C. K since it is above Aluminium in the reactivity series

Leviafan [203]3 years ago

5 0

C. K

Recall the activity series of metals. The order is

K > Al > Mn > Zn

Mn and Zn are below Al in the activity series.

K is the only metal that is above Al.

∴ K will replace Al in a single displacement reaction.

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What is the sum of 2.7 g and 2.47g expressed in the correct number of significant digits

NeTakaya

In order to get the sum of 2.7 and 2.47, we will add both numbers. So, 2.7 plus 2.47 would be 5.17. And the correct number of significant digits in this number is still the same, 5.17, so we have three significant numbers. Why? There are three rules in identifying the significant figures: 1. Non zeros are always significant (which this applies in our sum above). 2. Any zeros in between significant numbers are always significant. 3. The final zero or zeros in the decimal portion only are significant.

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3 years ago

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Why is this mars landing a new era for space exploration

Artemon [7]

Answer: The scientific reasons for going to Mars can be summarised by the search for life, understanding the surface and the planet's evolution, and preparing for future human exploration. Understanding whether life existed elsewhere in the Universe beyond Earth is a fundamental question of humankind.

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3 0

3 years ago

The decomposition of copper(II) nitrate on heating is endothermic reaction. 2Cu(NO3)2(s) → 2C10(s) + 4NO2(g) + O2(g) Calculate t

Basile [38]

Answer:

The enthalpy change for the given reaction is 424 kJ.

Explanation:

$2Cu(NO_3)_2(s)\rightarrow 2CuO(s) + 4NO_2(g) + O_2(g),\Delta H_{rxn}=?$

We have :

Enthalpy changes of formation of following s:

$\Delta H_{f,Cu(NO_3)_2}=-302.9 kJ/mol$

$\Delta H_{f,CuO}=-157.3 kJ/mol$

$\Delta H_{f,NO_2}= 33.2 kJ/mol$

$\Delta H_{f,O_2}= 0 kJ/mol$ (standard state)

$\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]$

The equation for the enthalpy change of the given reaction is:

$\Delta H_{rxn}$ =

$=(2 mol\times \Delta H_{f,CuO}+4\times \Delta H_{f,NO_2}+1 mol\times \Delta H_{f,O_2})-(2mol\times \Delta H_{f,Cu(NO_3)_2})$

$\Delta H_{rxn}$ =

$(2mol\times (-157.3 kJ/mol)+4\times 33.2 kJ/mol=1 mol\times 0 kJ/mol)-(2 mol\times (-302.9 kJ/mol)$

$\Delta H_{rxn}=424 kJ$

The enthalpy change for the given reaction is 424 kJ.

6 0

3 years ago

I need to check some chemistry questions. Help with any question is appreciated! :) Please include an explanation with your conc

finlep [7]

1. 4.67 kg; 2. 4.8 ×10^5 kg; 3. 0.106 cm^3; 4. 1.7 g/cm^3

Q1. Mass of Hg

Mass = 345 mL × (13.53 g/1 mL) = 4670 g = 4.67 kg

Q2. Mass of Pb

Step 1. Calculate the volume of the Pb.

V = lwh = 6.0 m × 3.5 m × 2.0 m = 42.0 m^3

Step 2. Calculate the mass of the Pb.

Mass = 42.0 m^3 × (11 340 kg/1 m^3) = 4.8 × 10^5 kg

Q3. Volume of displaced water

Volume of Ag = 0.987 g × (1 cm^3/9.320 g) = 0.106 cm^3

Archimedes: volume of displaced water = volume of Ag = 0.106 cm^3

4. Density of metal

Step 1. Convert ounces to grams

Mass = 3.35 oz × (28.35 g/1 oz) = 94.97 g

Step 2. Calculate the volume in cubic inches

V = lwh = 3.0 in × 2.5 in × 0.45 in = 3.38 in^3

Step 3. Convert cubic inches to cubic centimetres

V = 3.38 in^3 × (2.54 cm/1 in)^3 = 55.3 cm^3

Step 4. Calculate the density

ρ = m/V = (94.97 g/55.3 cm^3) = 1.7 g/cm^3 (magnesium?)

4 0

3 years ago

Type the correct answer in the box. Express your answer to two significant figures. An industrial vat contains 650 grams of soli

nikklg [1K]

88.98 %

The Balance Chemical Equation is as follow,

2 HCl + Pb(NO₃)₂ → 2 HNO₃ + PbCl₂

According to equation,

331.2 g (1 mole) Pb(NO₃)₂ produces = 278.1 g (1 mole) PbCl₂

So,

870 g of Pb(NO₃)₂ will produce = X g of PbCl₂

Solving for X,

X = (870 g × 278.1 g) ÷ 331.2 g

X = 730.5 g of PbCl₂

Therefore,

Theoretical Yield = 730.5 g

Also as given,

Actual Yield = 650 g

So using following formula for percentage yield,

%age Yield = (Actual Yield / Theoretical Yield) × 100

Putting values,

%age Yield = (650 g / 730.5 g) × 100

%age Yield = 88.98 %

Brianliest please and thank you.

6 0

3 years ago