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AleksandrR [38]
3 years ago
10

The temperature reading of –14°C corresponds to a Kelvin reading of:

Chemistry
1 answer:
Sedaia [141]3 years ago
3 0
The corrects answer is 259 K
You might be interested in
A chemist measures the energy change ?H during the following reaction: 2HgO (s) ?2Hg (l) +O2 (g) =?H182.kJ Use the information t
Leni [432]

Answer:

The reaction is endothermic.

Yes, absorbed

3.06x10¹kJ are absorbed

Explanation:

In the reaction:

2HgO(s) → 2Hg(l) + O₂(g) ΔH = 182kJ

As ΔH >0,

<em>The reaction is endothermic</em>

<em />

As the reaction is endothermic, when the reaction occurs,

<em>the heat is absorbed.</em>

<em></em>

Now, based on the equation, when 2 moles of HgO (Molar mass: 216.59g/mol), 182kJ are absorbed.

72.8g are:

72.8g * (1mol / 216.59g) = 0.3361 moles HgO.

that absorb:

0.3361 moles HgO * (182kJ / 2 moles) =

<h3>3.06x10¹kJ are absorbed</h3>
8 0
3 years ago
How much energy is required to raise the temperature of 10.7 grams of gaseous helium from 22.1 °C to 39.4 °C ?
Rainbow [258]

Answer:

Q = 2640.96 J

Explanation:

Given data:

Mass of He gas = 10.7 g

Initial temperature = 22.1°C

Final temperature = 39.4°C

Heat absorbed = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree. Specific heat capacity of He is 14.267 J/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 39.4°C - 22.1°C

ΔT = 17.3°C

Q = 10.7 g× 14.267 J/g.°C ×  17.3°C

Q = 2640.96 J

5 0
3 years ago
What volume in milliliters of 1.420 M sulfuric acid is needed to neutralize 3.209 g of aluminum hydroxide 3 H 2 SO 4 (aq)+2 Al(O
nadya68 [22]

Answer:

V=43.46mL

Explanation:

Hello!

In this case, since the reaction between sulfuric acid and aluminum hydroxide is:

3H_2SO_4+2Al(OH)_3\rightarrow Al_2(SO_4)_3+6H_2O

Whereas the ratio of sulfuric acid to aluminum hydroxide is 3:2; thus, we first compute the moles of sulfuric acid that complete react with 3.209 g of aluminum hydroxide:

n_{H_2SO_4}=3.209gAl(OH)_3*\frac{1molAl(OH)_3}{78.00gAl(OH)_3} *\frac{3molH_2SO_4}{2molAl(OH)_3} \\\\n_{H_2SO_4}=0.0617molH_2SO_4

Then, given the molarity, it is possible to obtain the milliliters as follows:

V=\frac{n}{M}=\frac{0.0617mol}{1.420mol/L}*\frac{1000mL}{1L}\\\\V=43.46mL

Best regards!

7 0
3 years ago
for the reactants represented by the equation 2H2 + O2 &gt; 2H2O, how many grams of water are produced from 376.97 mol of hydrog
Hunter-Best [27]

Answer:

6792.1 g

Explanation:

Dimensional analysis:

376.92 mol H2 * (2 mol h2o / 2 mol h2) * (18.02g h2o / 1 mol h2o)

6792.1 g

8 0
3 years ago
Read 2 more answers
Cover echo de cobalto
Alex17521 [72]

Answer: que?

Explanation:

7 0
3 years ago
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