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NeX [460]
4 years ago
9

I’m confused. Need any help.

Chemistry
1 answer:
Travka [436]4 years ago
4 0

Answer:

Problem 1 => 8778 joules

Problem 2 => 14,630 joules

Explanation:

Reference the Heating Curge of Water Problem posted earlier. These are temperature change fragments of that type problem. As you read the problem note the 'temperature change' phrase in the problem. This should signal you to use the q = m·c·ΔT expression as opposed to the q = m·ΔH expression where no temperature change is noted; i.e., melting or evaporation/boiling.

For the listed problems...

Problem 1: Amount of heat needed to heat 150 grams water from 21.0 to 35.0 Celcius.

Note the temperature change in the problem context => use q = m·c·ΔT ...

q = (150g)(4.18j/g·°C)(35.00°C - 21.0°C) = 8778 joules (4 sig. figs. based on the 150.0g value of water)

Problem 2: Amount of heat needed to heat 250.0 grams water from 31.0 to 45.0 Celcius.

Same type problem. Note temperature change in text of problem => use q = m·c·ΔT ...

q = (250.0g)(4.18j/g·°C)(45.0°C - 31.0°C) = 14,630 joules

Hope this helps. Doc

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Write a net ionic equation for the overall reaction that occurs when aqueous solutions of nitrous acid and sodium hydroxide are
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Answer:

H+(aq) +  OH-(aq) → H2O(l)

Explanation:

Step 1: Data given

nitrious acid = HNO3

sodium hydroxide = NaOH

Step 2: The unbalance equation

HNO3(aq) + NaOH(aq) →NaNO3(aq) + H2O(l)

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side (Ba^2+ and Br-), look like this:

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What is significant about the temperature absolute zero?
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Explanation:

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Determine the mass of CaCO3 required to produce 40.0 mL CO2 at STP. Hint use molar volume of an ideal gas (22.4 L)
cupoosta [38]

Answer:

m_{CaCO_3}=0.179gCaCO_3

Explanation:

Hello,

In this case, since the undergoing chemical reaction is:

CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:

PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2

Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:

m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3

Regards.

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