What happens when you mix hydrogen and oxygen?

Is carbon monoxide an element a compound a heterogeneous mixture or a homogeneous mixture?

Scrat [10]

Answer: Carbon monoxide, or "CO", is a compound. This is a homogeneous compound (not a mixture).
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3 0

3 years ago

A reaction is occurring in a test tube. How is heat transmitted to the surroundings?

Cerrena [4.2K]

It should be noted that when a reaction is occurring in a test tube, heat transmitted to the surroundings when Molecules collide with the glass, and the glass molecules then transmit that energy to the outside.

Heat can be regarded as a form of energy which is energy that is been transferred as a result of difference in temperature.

In the case above, Molecules collide with the glass, and the glass molecules then transmit that energy to the outside which is an exothermic reaction.

Therefore, option B is correct.

Learn more about heat at:

brainly.com/question/12072129

8 0

3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3.

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

4 years ago

How much would the temperature of 275 g of water increase if 36.5 kj of heat were added?

Andrew [12]

<u>Answer:</u> The temperature increase will be 31.70°C.

<u>Explanation:</u>

To calculate the increase in the temperature of the system, we use the equation:

$q=mc\Delta T$

where,

q = Heat absorbed = 36.5 kJ = 36500J

m = Mass of water = 275 g

c = Specific heat capacity of water = $4.186J/g^oC$

$\Delta T$ = change in temperature = ? °C

Putting values in above equation, we get:

$36500J=275\times 4.186J/g^oC\times \Delta T\\\\\Delta T=31.70^oC$

Hence, the temperature increase will be 31.70°C.

8 0

3 years ago

Read 2 more answers

electron affinity of lithium is -60 whereas of cesium is -45.this trend is due to... plz give me accurate answer

lesya692 [45]

Electron affinity is defined as the change in energy (in kJ/mole) of a neutral atom (in the gaseous phase) when an electron is added to the atom to form a negative ion. In other words, the neutral atom's likelihood of gaining an electron.

Electron Affinity of Lithium is 59.6 kJ/mol.

Electron Affinity of Caesium is 45.5 kJ/mol.

Electron Affinity of Lithium is 59.6 kJ/mol. Electronegativity of Lithium is 0.98. ... Electron affinities are more difficult to measure than ionization energies. An atom of Lithium in the gas phase, for example, gives off energy when it gains an electron to form an ion of Lithium.

Trends

The ionization energy of the elements within a period generally increases from left to right. This is due to valence shell stability.

The ionization energy of the elements within a group generally decreases from top to bottom. This is due to electron shielding.

The noble gases possess very high ionisation energies because of their full valence shells as indicated in the graph. Note that helium has the highest ionization energy of all the elements.

4 0

3 years ago