How many liters of nitrogen dioxide are in 16 g of NO2 at STP?

An exothermic reaction has a positive enthalpy (heat) of reaction.(T/F)

mestny [16]

Answer:

True.

Explanation:

An exothermic reaction has a positive enthalpy (heat) of reaction. However, it can be negative in some circumstances.

5 0

3 years ago

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What is the pH of a solution that contains 25 grams of hydrochloric acid (hcl) dissolved in 1.5 liters of water?

Darina [25.2K]

First we calculate the concentration of HCl:

Moles = mass / Mr
= 25 / 36.5
= 0.685 mol

Concentration = 0.685/1.5 = 0.457 mol / dm³

For a strong monoprotic acid, the concentration of hydrogen ions is equal to the acid concentration.

pH = -log[H+]
pH = -log(0.457)
= 0.34

4 0

3 years ago

When copper oxide is reacted with sulfuric acid, the acid is often gently heated. This is done in order to:

pav-90 [236]

Copper oxide(solid) + Sulphuric Acid (aqueous)-> Copper Sulphate (aqueous)+ Water(liquid)

In equation form:

CuO +H2SO4 -> CuSO4 + H2O

The colour change you will see is black to blue as Copper oxide is usually found as a black powder. Upon the reaction with sulphuric acid it will change to a cyan blue.

If you heat the made solution of copper sulphate, the water will evaporate and you will be left with white anhydrous copper sulphate crystals.

7 0

3 years ago

Two samples of the same compound are compared. what does the data represent? sample 1: 24.22 g carbon and 32.00 g oxygen sample

goblinko [34]

According to law of definite proportion:

In a compound, elements are always arranged in fixed ratio by mass.

Here, sample 1 has 23.22 g Carbon and 32.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

$n_{C}=\frac{m_{C}}{M_{C}}=\frac{24.22 g}{12 g/mol}\approx 2 mol$

Similarly, number of moles of oxygen will be:

$n_{O}=\frac{m_{O}}{M_{O}}=\frac{32 g}{16 g/mol}=2 mol$

The ratio of number of moles of carbon and oxygen will be:

$C:O=n_{C}:n_{O}=2:2=1:1$

Therefore, formula of compound will be CO.

Sample 2:

It has 36.22 g Carbon and 48.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

$n_{C}=\frac{m_{C}}{M_{C}}=\frac{36.22 g}{12 g/mol}\approx 3 mol$

Similarly, number of moles of oxygen will be:

$n_{O}=\frac{m_{O}}{M_{O}}=\frac{48 g}{16 g/mol}=3 mol$

The ratio of number of moles of carbon and oxygen will be:

$C:O=n_{C}:n_{O}=3:3=1:1$

The formula of compound will be CO.

Therefore, it is proved that carbon and oxygen are present in fixed ratios in both the samples.

4 0

3 years ago

How many grams of mercury can be produced if 18.0 g of mercury (11) oxide decomposes?

NARA [144]

Answer:

18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury

Explanation:

Mercury oxide has molar mass of 216.6 g/ mol. It gas a molecular formula of HgO.

The decomposition of mercury oxide is given by the chemical equation below:

2HgO ----> 2Hg + O₂

2 moles of HgO decomposes to produce 1 mole of Hg

2 moles of HgO has a mass of 433.2 g

433.2 g of HgO produces 216.6 g of Hg

18.0 of HgO will produce 18 × 216.6/433.2 g of Hg = 9.0 g of Hg

Therefore, 18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury

3 0

3 years ago