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3 years ago

2H2(g) + O2(g) mc001-1.jpg 2H2O(g)

Chemistry

2 answers:

Aleks [24]3 years ago

6 0

The answer is a) by increasing the temperature of the reactants.

Lesechka [4]3 years ago

6 0

Answer:

Explanation:

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Someone help me plz :))) brainliest

SOVA2 [1]

Answer:

Option 2

Explanation:

3 0

3 years ago

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago

Increased use of incineration is sometimes advocated as a safe way to dispose of chemical waste. But opponents of incineration p

pochemuha

Answer:

Option A is correct.

At the two incinerators at which leaks were reported, staff had had only cursory training on the proper procedures for incinerating chemical waste.

Explanation:

The main aim of the argument presented is to talk down the use of burning by incinerators method to dispose chemical waste. The argument presents great points in that there were 40 incidents at two existing commissioned incinerators in the last year where unexpected releases of dangerous chemical agents happened.

So, basically, the argument is all about how the high frequency of the unsuspected release of dangerous chemical agents should discourage this method of chemical waste disposal.

The argument then concluded that if more chemical waste are disposed using the burning by incinerator method, there will be more unsuspected release of dangerous chemicals.

We are then required to find the statement that most weakens the conclusion that there will be more toxic releases if more chemical waste are burned.

Analysing the Statements one by one

Statement A

This statement provides a possible reason for this high frequency of dangerous chemical releases. It states that the staff haven't been properly trained. So, this means that properly training the staff should most likely lead to lesser cases of toxic releases into the environment.

This is the statement that most weakens the conclusion.

Statement B

The conclusion wasn't about the incinerator method being the safest method. It was about whether increased incineration would lead to more toxic relaeses. So, this doesn't affect the conclusion.

Statement C

This statement says that incineration can be increased without building new incinerators by tapping into unused capacity at the old incinerators. Also doesn't affect rhe conclusion whether increased use of incineration will lead to more toxic leaks.

Statement D

This statement strengthens the argument; which is the opposite of what we're aiming to achieve.

Statement E

This statement hints that the toxic leaks do not have that much of a harmful effect because the toxic releases do not go beyond the property of the incinerator. This also doesnt tackle the conclusion about the frequency of leaks, it only addresses how not harmful the toxic leaks can be.

Hope this Helps!!!

5 0

3 years ago

A chemist is about to perform a titration by adding a base to an acid. what is the most important reason for using a base instea

zloy xaker [14]

The answer to this is to prevent dilution of the base, which may affect the results. I think that is right.

6 0

3 years ago

Read 2 more answers

You have a flask with 1.2 L of liquid in it, how many mL is this?

mihalych1998 [28]

I literally is 1000 ml. Therefor 1.2 literally is 1.2*1000 which is 1200 ml