Flame tests are used to identify the presence of a relatively small number of metal ions in a compound. Not all metal ions give flame colors. For Group 1 compounds, flame tests are usually by far the easiest way of identifying which metal you have got.
Answer:
ΔS > 0 only for choice E: CH4(g) + H2O (g) → CO(g) + 3 H2(g)
Explanation:
Our strategy in this question is to use the trend in entropies :
S (solids) less than S (liquids) less than S (gases)
Also we have to look for the molar quanties involved of each state and their change to answer the question:
A. N2(g) + 3 H2(g) → 2 NH3(g)
Here we have 4 moles gases going to 2 moles of products, so the change in entropy is negative.
B. Na2CO3(s) + H2O(g) + CO2(g) → 2 NaHCO3(s)
The change in entropy is negative since we have 2 mol gases in the reactants and zero in the products.
C. CH3OH(l) → CH3OH(s)
A liquid has a higher entropy than a solid so ΔS is negative
D. False see A,B,C
E. The change in moles of gases is 4 - 2= 2, therefore ΔS is greater than O.
Answer:
Carbon Dioxide = CO2
Explanation:
The synthesis of Malachite is seen in the chemical formula:
CuSO 4 . 5H2O(aq) + 2NaCO3(aq) --> CuCO 3 Cu(OH) 2 (s) + 2Na 2 SO 4 (aq) + CO 2 (g) + 9H 2 O(l)
The bubbles mentioned in the question hints that our interest is the compounds in their gseous phase (g).
Upon examining the chemical equation, only CO2 is in the gaseous state and hence the only one that can be formed as bubbles,
Answer:
79'420J
Explanation:
We know mass 250g and
∆T=98°C -22°C
=76°C
The specific heat capacity for water is 4,18Zj/g°C
THEREFORE
∆H=250g•14,8j/g°C •76°C
=79'420J
Answer:
KOH(aq) + HCI(aq) -----> KCI(aq )+ H2O
base acid salt water
hope this helps :)
Explanation: