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3 years ago

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Equal volumes of H2 and o2 are placed in a balloon and then ignited. Assuming that the reaction goes to completion, which gas wi

ll remain in excess 2 O 2H both are in equal proportion oxygen water hydrogen

2 answers:

Anastasy [175]3 years ago

7 0

You will have excess O2. The ideal gas law dictates that all other variables kept the same, equal volume means equal number of moles.

Naddika [18.5K]3 years ago

5 0

Answer : The correct option is, Oxygen.

Explanation :

The given balanced chemical reaction is:

$2H_2(g)+O_2(g)\rightarrow 2H_2O(g)$

From the balanced reaction we conclude that, 2 moles of hydrogen react with 1 mole of oxygen to give 2 moles of water.

As we know that at STP, 1 mole of gas contains 22.4 L volume of gas. That means there is a direct relation between the moles and volume of gas.

So we ca say that,

2 unit volume of hydrogen gas will react with 1 unit volume of oxygen gas.

As per question, the volume of gasses are equal that means half of oxygen gas will be consumed against the complete volume of hydrogen gas.

Thus, the gas that will remain in excess is oxygen gas.

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Two unknown molecular compounds were being studied. A solution containing 5.00 g of compound A in 100. g of water froze at a low

LenaWriter [7]

Answer:

Compound B has greater molar mass.

Explanation:

The depression in freezing point is given by ;

$\Delta T_f=i\times k_f\times m$ ..[1]

$m=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}$

Where:

i = van't Hoff factor

$k_f$ = Molal depression constant

m = molality of the solution

According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.

The depression in freezing point of solution with A solute: $\Delta T_{f,A}$

Molar mass of A = $M_A$

The depression in freezing point of solution with B solute: $\Delta T_{f,B}$

Molar mass of B = $M_B$

$\Delta T_{f,A}>\Delta T_{f,B}$

As we can see in [1] , that depression in freezing point is inversely related to molar mass of the solute.

$\Delta T_f\propto \frac{1}{\text{Molar mass of solute}}$

$M_A$

This means compound B has greater molar mass than compound A,

4 0

3 years ago

Consider the following reaction between mercury(II) chloride and oxalate ion.

Alina [70]

<u>Answer:</u> The rate law of the reaction is $\text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2$

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)$

Rate law expression for the reaction:

$\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b$

where,

a = order with respect to $HgCl_2$

b = order with respect to $C_2O_4^{2-}$

Expression for rate law for first observation:

$3.2\times 10^{-5}=k(0.164)^a(0.15)^b$ ....(1)

Expression for rate law for second observation:

$2.9\times 10^{-4}=k(0.164)^a(0.45)^b$ ....(2)

Expression for rate law for third observation:

$1.4\times 10^{-4}=k(0.082)^a(0.45)^b$ ....(3)

Expression for rate law for fourth observation:

$4.8\times 10^{-5}=k(0.246)^a(0.15)^b$ ....(4)

Dividing 2 from 1, we get:

$\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2$

Dividing 2 from 3, we get:

$\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2$

3 0

3 years ago

Please help,<br> How could you tell a Ca(NO3)2 solution from a Zn(NO3)2 solution?

tamaranim1 [39]

Answer:

you better give me brainliest

Explanation:

Zinc nitrate and calcium nitrate solution can be distinguished by reaction with ammonium hydroxide. Zinc forms a white gelatinous ppt. whereas there is no precipitation of calcium hydroxide even with excess of ammonium hydroxide

3 0

1 year ago

Can someone explain the steps for balancing chemical equations in depth?

m_a_m_a [10]

Answer:

Steps explained below

Explanation:

To explain balancing of chemical equations, I will make use of an example equation where Hydrogen and oxygen react to form water.

H2 + O2 = H2O

Now, the equation I've listed above is an unbalanced chemical equation. It can be balanced by the following steps;

Step 1: Identify the elements on both the left Hand side and the right hand side.

In this case;

on the left hand side, we have H and O.

On the right hand side, we have H and I also.

Step 2: Identify the number of atoms of each element on both the left and right hand sides.

On the left, H has 2 atoms and O has 2 atoms.

On the right, H has 2 atoms and O has 1 atom.

Step 3: For the equation to be balanced, the number of atoms of each element on the right and left hand side must be the same.

Thus,

O on the left hand side has 2 atoms but on the right hand side it has 1 atom. Thus, we will multiply O on the right by 2 to balance what we have on the left.

So, we now have;

H2 + O2 = 2H2O

Step 4: Check equation: We now have;

H2 + O2 = 2H2O

Our left hand side remains 2 atoms of H and 2 atoms of O. But on the right, we now have;

2 atoms O and 4 atoms of H.

Which means atoms of H is not balanced with the left side.

Step 5: rebalance equation: To rebalance, we multiply H on the left by 2 to give us 2 × 2 = 4 atoms.

Thus, we now have;

2H2 + O2 = 2H2O

3 0

3 years ago

Read 2 more answers

Rank the following compounds in order of decreasing acid strength using periodic trends.Rank the acids from strongest to weakest

Naddika [18.5K]

HBr>H2S>H2Se>BH3 so basically

1. HBr

2. H2S

3. H2Se

4. BH3

4 0

4 years ago

Read 2 more answers