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3 years ago

the answer choices for all are shifts to the right, shifts to the left, and no change I need help plsssss

Chemistry

1 answer:

GenaCL600 [577]3 years ago

8 0

Do you still need the answer for these ? if so , i have them

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Iron oxide and hydrogen react to form iron and water in the following chemical equation:

just olya [345]

Answer:

The answer to your question is 245.9 g of Fe₂O₃

Explanation:

Data

grams of Fe₂O₃ = ?

grams of Fe = 172 g

Balanced chemical reaction

Fe₂O₃ + 3H₂ ⇒ 2Fe + 3H₂O

Process

1.- Calculate the molar mass of of Iron Oxide and Iron

Fe₂O₃ = (2 x 56) + (3 x 16) = 112 + 48 = 160 g

2Fe = 2 x 56 = 112 g

2.- Use proportions and cross multiplication to solve this problem

160 g of Fe₂O₃ --------------- 112 g of Fe

x ---------------- 172 g of Fe

x = (172 x 160)/112

x = 27520/112

x = 245.9 g of Fe₂O₃

7 0

3 years ago

Any info on Neptunes spot? I need it for a project.

LUCKY_DIMON [66]

Here's one! those spots are actually storms:)

3 0

3 years ago

The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C

artcher [175]

Explanation:

The given data is as follows.

$P_{1}$ = 100 mm Hg or $\frac{100}{760}atm$ = 0.13157 atm

$T_{1}$ = $1080 ^{o}C$ = (1080 + 273) K = 1357 K

$T_{2}$ = $1220 ^{o}C$ = (1220 + 273) K = 1493 K

$P_{2}$ = 600 mm Hg or $\frac{600}{760}atm$ = 0.7895 atm

R = 8.314 J/K mol

According to Clasius-Clapeyron equation,

$log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}$

$log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]$

$log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]$

0.77815 = $\frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K$

$\Delta H_{vap}$ = $2.219 \times 10^{5}$ J/mol

= $2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}$

= 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

4 0

3 years ago

A corpse discovered in the desert at 5:00 PM was found to have larvae from the blow fly species Phormia regina that had just dev

horrorfan [7]

Answer:

12.19 hours prior to discovery was the corpse placed in the desert.

Explanation:

Heat required for fly eggs to develop into first instar larvae = $Exposure\,time\times temperature$

From the given,

Exposure time = 16 hrs

Temperature = 27 C

$=16\times 27 = 427.2^{o}C$

The day time exposed hours is 10 hrs.-(7.00am-5.00pm)

$=10\times 37.8 = 378^{o}C$

So, the extra heat = 427.2-378= 49.2

The addition heat divided by average temperature.

Average temperature = 22.4 C

$=\frac{49.2}{22.4}= 2.19 hrs$

So, the total time exposed is night time hours to day time hours.

= 10+2.19 = 12.19 hrs.

Therefore, 12.19 hours prior to discovery was the corpse placed in the desert.

3 0

3 years ago

What is the total pressure of a gas mixture containing partial pressures of

Advocard [28]

Answer:

1.54 atm

Explanation:

By Dalton's Law Of partial pressure,

Total Pressure = Sum of all partial pressures

So,P= P1 + P2 + P3

Therefore, P=0.23+0.42+0.89

=1.54 atm

8 0

2 years ago