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musickatia [10]
3 years ago
10

An element is a ________ substance A. Mixed B. Pure C. Combined D. Human-made

Chemistry
2 answers:
Alex787 [66]3 years ago
8 0

Answer:

An element is a pure substance made up of identical atoms.

Explanation:

hope it helps

and brainliest if helped......

harina [27]3 years ago
3 0

Answer:

b, pure.

Explanation:

an element is a pure substance and it cannot be broken down hope this helped :)

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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
Which pair shares the same empirical formula?
ZanzabumX [31]

Answer:- 3. CH_3 and C_2H_6

Explanations:- An empirical formula is the simplest whole number ratio of atoms of each element present in the molecule/compound.

For example, the molecular formula of benzene is C_6H_6 . The ratio of C to H in it is 6:6 that could be simplified to 1:1. So, an empirical formula of benzene is CH.

In the first pair, the ratio of C to H in first molecule is 2:4 that could be simplified to 1:2 and  the empirical formula is CH_2 . In second molecule the ratio of C to H is 6:6 and it could be simplified to 1:1. and the empirical formula is CH. Empirical formulas are different for both the molecules of first pair and so it is not the right choice.

In second pair, C to H ratio in first molecule is 1:2, so the empirical formula is CH_2 . The C to H ratio for second molecule is 1:4, so the empirical formula is CH_4 . Here also, the empirical formulas are not same and hence it is also not the right choice.

In third pair, C to H ratio in first molecule is 1:3, so the empirical formula is CH_3 . In second molecule the C to H ratio is 2:6 and it is simplified to 1:3. So, the empirical formula for this one is also CH_3 . Hence. this is the correct choice.

In fourth pair, first molecule empirical formula is CH. Second molecule has 2:4 that is 1:2 mole ratio of C to H and so its empirical formula is CH_2 . As the empirical formulas are different, it is not the right choice.

So, the only and only correct pair is the third one. 3. CH_3 and C_2H_6

4 0
3 years ago
Read 2 more answers
An alloy is a mixture that has
defon

Answer:

an alloy is a mixture of elements that has metallic properties

Explanation:

8 0
3 years ago
Read 2 more answers
Consider the reaction cacn2 + 3 h2o → caco3 + 2 nh3 . how much caco3 is produced if 47.5 moles nh3 are produced? 1. 0.950 g 2. 9
zhannawk [14.2K]
The  CaCO3  produced  if  47.5   moles of  NH3  produced  is   calculated  as  follows

CaCN2  +3H2O  = CaCO3  + 2NH3

by  use   of  mole  ratio  between  CaCO3 to NH3  which  is 1:2  the  moles  of  CaCO3  is therefore =  47.5 /2= 23.75  moles

mass of CaCO3   is  therefore = moles  x  molar  mass
= 23.75 moles  x  100g/mol=  2375 grams   which  is approximate   2380  grams(answer 6)
8 0
3 years ago
What's an example of chemical reaction
sergeinik [125]
S + O = SO4. cracked also chemical reaction
6 0
3 years ago
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