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nikitadnepr [17]
3 years ago
10

What's the ionic compound formula for zinc bicarbonate

Chemistry
2 answers:
Leya [2.2K]3 years ago
8 0
<span>Zn(HCO3)2
Zn: zinc
Carbonate : HCO3
and bi means we have 2 bicarbonate</span>
raketka [301]3 years ago
6 0
Zn(HCO3)2 is the compound for Zinc Bicarbonate
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4 0
4 years ago
What is the difference between liquid nitrogen and nitrogen gas​
Studentka2010 [4]

Answer:

The difference between liquid nitrogen and nitrogen gas is liquid nitrogen gas is man made or caused by humans but, nitrogen gas takes place naturally in the earths atmosphere.

Explanation:

Hope this helps :)

8 0
3 years ago
What is the MOLAR heat of combustion of methane(CH₄) if 64.00g of methane are burned to heat 75.0 ml of water from 25.00°C to 95
melamori03 [73]

Answer:

-5.51 kJ/mol

Explanation:

Step 1: Calculate the heat required to heat the water.

We use the following expression.

Q = c \times m \times \Delta T

where,

  • c: specific heat capacity
  • m: mass
  • ΔT: change in the temperature

The average density of water is 1 g/mL, so 75.0 mL ≅ 75.0 g.

Q = 4.184J/g.\°C \times 75.0g \times (95.00\°C - 25.00\°C) = 2.20 \times 10^{3} J = 2.20 kJ

Step 2: Calculate the heat released by the methane

According to the law of conservation of energy, the sum of the heat released by the combustion of methane (Qc) and the heat absorbed by the water (Qw) is zero

Qc + Qw = 0

Qc = -Qw = -22.0 kJ

Step 3: Calculate the molar heat of combustion of methane.

The molar mass of methane is 16.04 g/mol. We use this data to find the molar heat of combustion of methane, considering that 22.0 kJ are released by the combustion of 64.00 g of methane.

\frac{-22.0kJ}{64.00g} \times \frac{16.04g}{mol} = -5.51 kJ/mol

8 0
3 years ago
In the following reaction, which substance is the precipitate? (NH4)2SO4(aq) + Ba(NO3)2(aq) BaSO4(s) + 2NH4NO3(aq
mariarad [96]
Assuming that the reactants are:

(NH4)2SO4 (aq) + Ba(NO3)2 (aq)

and the products are:

BaSO4 (s) + 2NH4NO3 (aq),

then you will have to determine which product is insoluble. You should have access to solubility rules to help you determine this.

According to the solubility rules, the following elements are considered insoluble when paired with SO4:

Sr^2+, Ba^2+, Pb^2+, Ag^2+, and Ca^2+

Therefore, the precipitate will be BaSO4 (s).

5 0
3 years ago
Read 2 more answers
QuiCk AsAp HeLp mE FasT
zzz [600]

Answer:

b

Explanation:

6 0
3 years ago
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