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Margarita [4]
3 years ago
6

Using this illustration, what is the chemical formula? Picture and possible answers are below.

Chemistry
2 answers:
qwelly [4]3 years ago
4 0
NH3. 1 nitrogen and 3 hydrogen
aleksandrvk [35]3 years ago
3 0

The answer is It is NH3.....

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A sample of an unknown metal has a mass of 120.4 g. As the sample cools from 90.5°C to 25.7°C, it releases 7020 J of energy. Usi
Dafna1 [17]

Answer:

Explanation:

sheesh i wouldn’t know

6 0
2 years ago
How many grams of NaCI per kilogram of water are present in a 2.7 molal aqueous solution​
navik [9.2K]

Answer:

157.79 g

Explanation:

The definition of molality is:

  • molality = moles of solute / kilogram of solvent

This means that in a 2.7 molal solution, there are 2.7 moles of NaCl per kilogram of water.

So now w<u>e convert those 2.7 moles of NaCl to grams</u>, using its <em>molar mass</em>:

  • 2.7 mol * 58.44 g/mol = 157.79 g
5 0
3 years ago
An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at
lana [24]

Answer:

ΔU = −55.45 kJ

Explanation:

From first law of thermodynamics in chemistry, we have;

ΔU = Q + W

where;

ΔU is change in internal energy

Q is the net heat transfer

W is the net work done

We are given;

Q = 74.6 kJ

But Q will be negative since heat is released

Thus;

ΔU = -74.6 kJ + W

We are given;

Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²

Volume before reaction; Vi = 8.2 L = 0.0082 m³

Volume after reaction; V_f = 2.8 L = 0.0028 m³

Now,

W = -P(V_f - V_i)

W = - 3546375(0.0028 - 0.0082)

W = 19.15 KJ

Thus;

ΔU = Q + W

ΔU = -74.6 kJ + 19.15 KJ =

ΔU = −55.45 kJ

6 0
3 years ago
Write the net ionic equation for the precipitation of cobalt(ii) carbonate from aqueous solution
andrey2020 [161]
Missing question: Write the net ionic equation for the precipitation reaction that occurs when aqueous solutions of ammonium carbonate and cobalt(II) bromide are combined.<span>Balanced chemical reaction:
(NH</span>₄)₂CO₃(aq) + CoBr₂(aq) → CoCO₃(s) + 2NH₄Br(aq).
Net ionic reaction:
2NH₄⁺(aq) + CO²⁻(aq) + Co²⁺(aq) + 2Br⁻(aq) → CoCO₃ + 2NH₄(aq)+ 2Br(aq).
or CO²⁻(aq) + Co²⁺(aq) → CoCO₃(s).
4 0
3 years ago
How many milliliters of 0.0896M LiOH are required to titrate 25.0 mL of 0.0759M HBr to the equivalence point?
slavikrds [6]

Answer:

V_{LiOH}=21.8mL

Explanation:

Hello,

In this case, during titration at the equivalence point, we find that the moles of the base equals the moles of the acid:

n_{LiOH}=n_{HBr}

That it terms of molarities and volumes we have:

M_{LiOH}V_{LiOH}=M_{HBr}V_{HBr}

Next, solving for the volume of lithium hydroxide we obtain:

V_{LiOH}=\frac{M_{HBr}V_{HBr}}{M_{LiOH}} =\frac{0.0759M*25.0mL}{0.0896M} \\\\V_{LiOH}=21.8mL

Best regards.

8 0
2 years ago
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