Limes and lemons have a ph of 2 and are _____. acidic basic bases hydroxides

Calculate the energy required to heat 322.0g of ethanol from −2.2°C to 19.6°C . Assume the specific heat capacity of ethanol und

just olya [345]

Answer:

There is 17.1 kJ energy required

Explanation:

Step 1: Data given

Mass of ethanol = 322.0 grams

Initial temperature = -2.2 °C = 273.15 -2.2 = 270.95K

Final temperature = 19.6 °C = 273.15 + 19.6 = 292.75 K

Specific heat capacity = 2.44 J/g*K

Step 2: Calculate energy

Q = m*c*ΔT

⇒ m = the mass of ethanol= 322 grams

⇒ c = the specific heat capacity of ethanol = 2.44 J/g*K

⇒ ΔT = T2 - T1 = 292.75 - 270.95 = 21.8 K

Q = 322 * 2.44 * 21.8 = 17127.8 J = 17.1 kJ

There is 17.1 kJ energy required

3 0

3 years ago

An unknown amount of mercury (II) oxide was decomposed in the lab. Mercury metal was formed and 5.20 L of oxygen was released at

Yakvenalex [24]

Answer:

68.3g

Explanation:

1. Word equation:

mercury(II) oxide → mercury + oxygen

2. Balanced molecular equation:

2HgO → 2Hg + O₂(g)

3. Mole ratio

Write the ratio of the coefficients of the substances that are object of the problem:

$2molHgO/1molO_2$

4. Calculate the number of moles of O₂(g)

Use the equation for ideal gases:

$pV=nRT\\\\\\n=\dfrac{pV}{RT}\\\\\\n=\dfrac{0.970atm\times5.20L}{0.08206atm.L/K.mol\times 390.0K}\\\\\\n=0.1576mol$

5. Calculate the number of moles of HgO

$\dfrac{2molHgO}{1molO_2}\times 0.1576molO_2=0.315molHgO$

6. Convert to mass

mass = # moles × molar mass

molar mass of HgO: 216.591g/mol

mass = 0.315mol × 216.591g/mol = 68.3g

7 0

3 years ago

Given the following data:N2(g) + O2(g)→ 2NO(g), ΔH=+180.7kJ2NO(g) + O2(g)→ 2NO2(g), ΔH=−113.1kJ2N2O(g) → 2N2(g) + O2(g), ΔH=−163

statuscvo [17]

Answer:

ΔH = +155.6 kJ

Explanation:

The Hess' Law states that the enthalpy of the overall reaction is the sum of the enthalpy of the step reactions. To do the addition of the reaction, we first must reorganize them, to disappear with the intermediaries (substances that are not presented in the overall reaction).

If the reaction is inverted, the signal of the enthalpy changes, and if its multiplied by a constant, the enthalpy must be multiplied by the same constant. Thus:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

2NO(g) + O₂(g) → 2NO₂(g) ΔH = -113.1 kJ

2N₂O(g) → 2N₂(g) + O₂(g) ΔH = -163.2 kJ

The intermediares are N₂ and O₂, thus, reorganizing the reactions:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

NO₂(g) → NO(g) + (1/2)O₂(g) ΔH = +56.55 kJ (inverted and multiplied by 1/2)

N₂O(g) → N₂(g) + (1/2)O₂(g) ΔH = -81.6 kJ (multiplied by 1/2)

------------------------------------------------------------------------------------

N₂O(g) + NO₂(g) → 3NO(g)

ΔH = +180.7 + 56.55 - 81.6

ΔH = +155.6 kJ

5 0

3 years ago

For which of the following activities might you want to hire a chemist?

Basile [38]

Answer:

OPTION (A) : Testing a rock sample for gold content

Explanation:

For testing a rock sample of gold content you will need a Chemist. To test the material, the sample is rubbed on black stone which will leave a mark on the stone. This mark is tested by applying aqua fortis i.e nitric acid on the mark. If the mark gets dissolve then the material is not gold. If the mark sustain the it is further tested by applying aqua regia i.e nitric acid and hydrochloric acid which will prove the sample is of gold if it gets dissolve on using hydrochloric acid. The purity of the sample can be checked by differing the concentration of the aqua regia and comparing it with the gold material of the known purity.

7 0

3 years ago

Read 2 more answers

If the length of a string increases 2 times but the mass of the string remains constant, the new density of the string will equa

Genrish500 [490]

If the length of a string increases 2 times but the mass of the string remains constant, the original density will be multiplied by a factor mathematically given as

v=1.414 times

<h3>Will the new density of the string equal the original density multiplied by what factor?</h3>

Generally, the equation for the volume is mathematically given as

v = sqt(TL/m)

Where

density = mass/length

Therefore

$v2/v1 = \sqrt{L2/L1}$

$v2/v1 = \sqrt{2}$

In conclusion,v ratio will give us

v=1.414 times