Which element is present in all organic

What is the specific heat of ethanol if 4650 j are required to raise the temperature of 250.0g of ethanol from 22.0oc to 29.6 de

Alchen [17]

Answer:

$C=2.45\frac{J}{g\°C}$

Explanation:

Hello there!

In this case, since the thermodynamic definition of heat in terms of mass, specific heat and temperatures is given by:

$Q=mC(T_2-T_1)$

We are to calculate the specific heat of the ethanol as shown below:

$C=\frac{Q}{m(T_2-T_1)}$

Thus, by plugging it the given data we can obtain:

$C=\frac{4650J}{250.0g(29.6-22.00\°C)} \\\\C=2.45\frac{J}{g\°C}$

Regards!

5 0

3 years ago

CAN SOMEONE HELP ME ON THIS PLEASE

lord [1]

option C gas.............................. ᕕ( ᐛ )ᕗ

6 0

3 years ago

Define aqua regia short answer boreeeeeee

larisa [96]

Answer:

Aqua regia is a mixture of nitric acid and hydrochloric acid, optimally in a molar ratio of 1:3. Aqua regia is a yellow-orange fuming liquid, so named by alchemists because it can dissolve the noble metals gold and platinum, though not all metals.

Explanation:

3 0

3 years ago

Read 2 more answers

The balanced combustion reaction for C 6 H 6 is 2 C 6 H 6 ( l ) + 15 O 2 ( g ) ⟶ 12 CO 2 ( g ) + 6 H 2 O ( l ) + 6542 kJ If 8.10

Nataly [62]

Answer:

The final temperature of water = 35.2 °C

Explanation:

Step 1: Data given

Mass of C6H6 = 8.100 grams

Mass of water = 5691 grams

Temperature = 21 °C

Step 2: The balanced equation

2C6H6(l) + 15O2(g) ⟶ 12 CO2 (g) + 6H2O (l) + 6542 kJ

Step 3:

Q = m*c*ΔT.

⇒with Q = the heat released during this reaction (this depends on the amount of reactants used)

⇒ with m= the mass of the water

⇒with c = the "specific heat" of water = how much energy it takes to raise the temp of 1g of water by 1°C

⇒with ΔT = the change in the temperature of the water

For every 2 moles of C6H6 consumed, 6542 kJ of heat is released.

Step 4: Calculate moles for 8.100 grams

8.100grams / 78.11 g/mol= 0.1037 mol es

So, according to the equation, the amount of heat released is:

(0.1037 moles / 2 moles)* (6542 kJ) = 339.2 kJ

Step 5: Calculate the final temperature

Q = mcΔT

ΔT = Q / (m*c)

T2- T1 = Q / (m*c)

T2 = [Q / (m*c)] + T1 = [(339.2 kJ) / (5691g)(0.004186 kJ/g°C)] + 21°C = 35.2°C

The final temperature of water = 35.2 °C

8 0

3 years ago

Three mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,

8_murik_8 [283]

Answer:

(a). 46,666.7 g/mol; 78,571.4 g/mol

(b). 86950g/mol; 46,666.7 g/mol.

(c). 86950g/mol; 43,333.33 g/mol

Explanation:

So, we are given the molar masses for the three samples as: 10,000, 30,000 and 100,000 g mol−1.

Thus, the equal number of molecule in each sample = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

The average molar mass = [ ( 10,000)^2 + (30,000)^2 + 100,000)^2] ÷ 10,000 + 30,000 + 100,000 = 78,571. 4 g/mol.

(b). The equal masses of each sample = 3/[ ( 1/ 10,000) + (1/30,000 ) + (1/100,000) ] = 20930.23 g/mol.

Average molar mass = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

(c). Equal masses of the two samples = (0.145 × 10,000) + (0.855 × 100,000)/ 0.145 + 0.855 = 86950g/mol.

The weight average molar mass = 1.7 + 10,000 + 100,000/ 1.7 + 1 = 43,333.33 g/mol.

6 0

3 years ago