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3 years ago

Chemical equations must be balanced to satisfy ...?

Chemistry

2 answers:

Gelneren [198K]3 years ago

6 0

Answer:

The law of conservation of matter

Explanation:

The answer is correct because I had the same question in my studies

Can I get brainliest please

murzikaleks [220]3 years ago

4 0

Answer:

the law of conservation of matter.

Explanation:

I am in chem

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Write the following number in Standard (Numeric) Form. 5.8x10-5

DerKrebs [107]

Answer:

the answer of this question is 0.000058

4 0

2 years ago

The combustion of a sample of butane, C4H10 (lighter fluid), produced 2.46 grams of water.

avanturin [10]

a. 0.137

b. 0.0274

c. 1.5892 g

d. 0.1781

e. 5.6992 g

<h3>Further explanation</h3>

Given

Reaction

2 C4H10 + 13O2 -------> 8CO2 + 10H2O

2.46 g of water

Required

moles and mass

Solution

a. moles of water :

2.46 g : 18 g/mol = 0.137

b. moles of butane :

= 2/10 x mol water

= 2/10 x 0.137

= 0.0274

c. mass of butane :

= 0.0274 x 58 g/mol

= 1.5892 g

d. moles of oxygen :

= 13/2 x mol butane

= 13/2 x 0.0274

= 0.1781

e. mass of oxygen :

= 0.1781 x 32 g/mol

= 5.6992 g

6 0

3 years ago

According to the second law of thermodynamics, heat energy released by an organism is _____. gone forever used by another animal

stealth61 [152]

According to the second law of thermodynamics, heat energy released by an organism is released into the <span>environment and unusable. This energy is also known as entropy which is defined as the tendency of things to be dispersed and spontaneous. This means that entropy is always random and becomes unavailable energy.</span>

4 0

3 years ago

Read 2 more answers

A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som

irinina [24]

Answer:

$m_{KBr}=6.030gKBr$

Explanation:

Hello.

In this case, since the normal boiling point of X is 117.80 °C, the boiling point elevation constant is 1.48 °C*kg*mol⁻¹, the mass of X is 100 g and the boiling point of the mixture of X and KBr boils at 119.3 °C, we can use the following formula:

$(T_b-T_b_0)=i*m*K_b$

Whereas the Van't Hoff factor of KBr is 2 as it dissociates into potassium cations and bromide ions; it means that we can compute the molality of the solution:

$m=\frac{T_b-T_b_0}{i*K_b}=\frac{(119.3-117.8)\°C}{2*1.48\°C*kg*mol^{-1}}\\ \\m=0.507mol/kg$

Next, given the mass of solventin kg (0.1 kg from 100 g), we compute the moles KBr:

$n_{KBr}=0.507mol/kg*0.1kg=0.0507mol$

Finally, considering the molar mass of KBr (119 g/mol) we compute the mass that was dissolved:

$m_{KBr}=0.0507mol*\frac{119g}{1mol} \\\\m_{KBr}=6.030gKBr$

Best regards.

3 0

3 years ago

A mixture containing nitrogen, hydrogen, and iodine established the following equilibrium at 400 °C:2NH3(g) + 3I2(g) ⇌ N2(g) + 6

miss Akunina [59]

Answer: The value of $K_{c}$ for this reaction is 250000.

Explanation:

The given equation is as follows.

$2NH_{3}(g) + 3I_{2}(g) \rightleftharpoons N_{2}(g) + 6HI(g)$

$N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g); K_{c_{1}} = 0.50$ ... (1)

$H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g); K_{c_{2}} = 50$ ... (2)

To balance the atoms, multiply equation (2) by 3. Hence, the equation (2) can be re-written as follows.

$3H_{2}(g) + 3I_{2}(g) \rightleftharpoons 6HI(g); K_{c_{2}} = (50)^{3}$ ... (3)

Now, subtract equation (1) from equation (3). So, the equation formed will be as follows.

$3I_{2} - N_{2} \rightleftharpoons 6HI - 2NH_{3}$

This equation can also be re-written as follows.

$3I_{2} + 2NH_{3} \rightleftharpoons N_{2} + 6HI$

This equation is similar to the equilibrium equation given to us.

Therefore, during this subtraction the equation constants get divided as follows.

$K^{'}_{c} = \frac{K_{c_{2}}}{K_{c_{1}}}\\= \frac{(50)^{3}}{0.50}\\= 250000$

Thus, we can conclude that the value of $K_{c}$ for this reaction is 250000.

6 0

2 years ago