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2 years ago

Simple chem question - please help!

1 answer:

7 0

Molar mass O₂ = 32.0 g/mol and N₂ = 28.0 g/mol

N₂ +2 O₂ → 2 NO₂

28.0 g -------> 32.0 * 2
7 g ----------> ?

Mass of O₂ = 7 * 32.0 * 2 / 28.0

Mass of O₂ = 448 / 28.0

= 16 g of O₂

Answer B

hope this helps!

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Dry ice is solid carbon dioxide. A 0.050-g sample of dry ice is placed in an evacuated 4.6-L vessel at 30 °C. Calculate the pres

goldenfox [79]

The answer is 6.1*10^-3 atm.

The pictures and explanations are there.

3 0

2 years ago

(45 pts) What is the theoretical yield (in g) of iron(III) carbonate that can be produced from 1.72 g of iron(III) nitrate and a

mote1985 [20]

Answer:

1.04g of iron III carbonate

Explanation:

First, we must put down the equation of reaction because it must guide our work.

2Fe(NO3)3(aq) + 3Na2CO3(aq)→Fe2(CO3)3(s) + 6NaNO3(aq)

From the question, we can see that sodium carbonate is in excess while sodium nitrate is the limiting reactant.

Number of moles of iron III nitrate= mass of iron III nitrate reacted/ molar mass of iron III nitrate

Mass of iron III nitrate reacted= 1.72g

Molar mass of iron III nitrate= 241.88 g∙mol–1

Number of moles of iron III nitrate= 1.72g/241.88 g∙mol–1= 7.11×10^-3 moles

From the equation of the reaction;

2 moles of iron III nitrate yields 1 mole of iron III carbonate

7.11×10^-3 moles moles of iron III nitrate yields 7.11×10^-3 × 1/ 2= 3.56×10^-3 moles of iron III carbonate

Theoretical mass yield of iron III carbonate = number of moles of iron III carbonate × molar mass

Theoretical mass yield of iron III carbonate = 3.56×10^-3 moles ×291.73 g∙mol–1 = 1.04g of iron III carbonate

8 0

2 years ago

Read 2 more answers

Calculate the specific heat capacity for a 22.7-g sample of lead that absorbs 237 J when its temperature increases from 29.8 °C

soldier1979 [14.2K]

Answer:

$\boxed {\boxed {\sf c\approx 0.159 \ J/ g \textdegree C}}$

Explanation:

We are asked to find the specific heat capacity of a sample of lead. The formula for calculating the specific heat capacity is:

$c= \frac{Q}{m \times \Delta T}$

The heat absorbed (Q) is 237 Joules. The mass of the lead sample (m) is 22.7 grams. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature. The temperature increases from 29.8 °C to 95.6 °C.

ΔT = final temperature -inital temperature
ΔT= 95.6 °C - 29.8 °C = 65.8 °C

Now we know all three variables and can substitute them into the formula.

Q= 237 J
m= 22.7 g
ΔT = 65.8 °C

$c= \frac {237 \ J}{22.7 \ g \ \times \ 65.8 \textdegree C}$

Solve the denominator.

22.7 g * 65.8 °C = 1493.66 g °C

$c= \frac {237 \ J}{1493.66 \ g \textdegree C}$

Divide.

$c= 0.1586706479 J /g \textdegree C$

The original values of heat, temperature, and mass all have 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place tells us to round the 8 up to a 9.

$c \approx 0.159 \ J/g \textdegree C$