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baherus [9]
2 years ago
7

Help please #7 and #8

Chemistry
1 answer:
ira [324]2 years ago
6 0
7 A lead
B Gold
Csilver

8 Afeathrr
B water
C silver
You might be interested in
How many significant figures<br> are in this number?<br> 43.55
leva [86]

Answer:

4 significant figures

Explanation:

Significant figures are the units/digits within a number that make the number more accurate and precise.

All digits (except for 0) are always significant. Therefore, all the digits in 43.55 are significant. Since there are 4 digits in the given number, there are 4 significant figures.

7 0
1 year ago
Reaction of hypochlorous acid and ammonia is product-favored system at equilibrium. Answer A: Reaction of hypochlorous acid and
pickupchik [31]

Answer:

A. True

B. False

Explanation:

Reaction between hypochlorous acid and ammonia does not yields a Ph value of 7.00 , this value is defined to water and no other reaction can yield exactly this Ph value. The acid reaction will give a Ph value below 7 while a base reaction will give Ph value of above 7.

7 0
2 years ago
Write the names and symbols for four elements in each of the following categories: (a) nonmetal, (b) metal, (c)metalloid.
Over [174]

Explanation:

Non-metals are the species that are electron deficient and they are able to accept one or more electrons from a donor atom in order to complete their octet.

For  example, carbon (C), nitrogen (N), chlorine, (Cl), phosphorus (P) etc are all non-metals.

Metals are the species that contain more number of electrons in their valence shell and in order to attain stability they easily lose an electron.

For example, sodium (Na), lithium (Li), Beryllium (Be), Magnesium (Mg) etc are all metals.

Metalloids are the species that show properties of both metals and non-metals.

For example, Boron (B), Antimony (Sb), Silicon (Si) and Germanium (Ge) etc are metalloids.

5 0
3 years ago
The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions
mamaluj [8]

Answer:

a) 62.1 kJ/mol

b) 2.82 kJ/mol

c) 270.91 kJ/mol

d) -851.5 kJ/mol

Explanation:

The enthalpy change for a reaction in standard conditions (ΔH°rxn) can be calculated by:

ΔH°rxn = ∑n*ΔH°f, products - ∑n*ΔH°f, reagents

Where n is the number of moles in the stoichiometry reaction, and ΔH°f is the enthalpy of formation at standard conditions. ΔH°f = 0 for substances formed by only a single element. The values can be found in thermodynamics tables.

a) 2Ag₂O(s) → 4Ag(s) + O₂(g)

ΔH°f, Ag₂O(s) = -31.05 kJ/mol

ΔH°rxn = 0 - (2*(-31.05)) = 62.1 kJ/mol

b) SnO(s) + CO(g) → Sn(s) + CO₂(g)

ΔH°f,SnO(s) = -285.8 kJ/mol

ΔH°f,CO(g) = -110.53 kJ/mol

ΔH°f,CO₂(g) = -393.51 kJ/mol

ΔH°rxn = [-393.51] - [-110.53 - 285.8] = 2.82 kJ/mol

c) Cr₂O₃(s) + 3H₂(g) → 2Cr(s) + 3H₂O(l)

ΔH°f,Cr₂O₃(s) = -1128.4 kJ/mol

ΔH°f,H₂O(l) = -285.83 kJ/mol

ΔH°rxn = [3*(-285.83)] - [( -1128.4)] = 270.91 kJ/mol

d) 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

ΔH°f,Fe₂O₃(s) = -824.2 kJ/mol

ΔH°f,Al₂O₃(s) = -1675.7 kJ/mol

ΔH°rxn = [-1675.7] - [-824.2] = -851.5 kJ/mol

3 0
3 years ago
I need help with this and quickyly if anyone is good in Chemistry??
Dominik [7]
The correct answer is d) chrima
6 0
3 years ago
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