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3 years ago

Can anyone help with properties of non-metals?

1 answer:

3 0

Dull (not shiny)
poor conductors of heat and electricity (they are insulators)
weak and brittle (they easily break or shatter when solid)

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Find the number of CoCl2 units present in a 0.78 mol sample.

stich3 [128]

$\qquad\qquad\huge\underline{{\sf Answer}}$

Here we go ~

1 mole of $\sf{ COCl_2}$ has 6.022 × 10²³ molecules of the given compound.

So, 0.78 mole of $\sf{COCl_2}$ will have ~

$\qquad \sf \dashrightarrow \: 0.78 \times 6.022 \times 10 {}^{23}$

$\qquad \sf \dashrightarrow \: \approx4.7 \times 10 {}^{23} \: \: \: molecules$

5 0

2 years ago

A fusion reaction releases energy because the binding energy of the resulting nucleus:______.

KIM [24]

Answer:

a. is released in the process

Explanation:

In fusion reaction the nucleus is unstable so it releases its binding energy resulting in decreasing its mass so it becomes more stable.

7 0

3 years ago

After balancing the following reaction under acidic conditions, how many mole equivalents of water are required and on which sid

nordsb [41]

Answer:

d. 8 moles of H2O on the product side

Explanation:

Hello,

In this case, we need to balance the given redox reaction in acidic media as shown below:

$MnO_4^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O\\\\2Cl^{1-}\rightarrow Cl_2^0+2e^-\\\\2*[(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O]\\\\5*[2Cl^{1-}\rightarrow Cl_2^0+2e^-]\\\\\\\\2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10e^- \rightarrow 2Mn^{2+}+8H_2O\\\\10Cl^{1-}\rightarrow 5Cl_2^0+10e^-\\$

Then, we add the half reactions:

$2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10Cl^{1-} \rightarrow 2Mn^{2+}+8H_2O+5Cl_2^0$

Thereby, we can see d. 8 moles of H2O on the product side.

Best regards.

4 0

3 years ago

An aqueous solution of calcium hydroxide is standardized by titration with a 0.112 M solution of hydrobromic acid. If 15.2 mL of

TiliK225 [7]

Answer:

0.0457 M

Explanation:

The reaction that takes place is:

2HBr + Ca(OH)₂ → CaBr₂ + 2H₂O

First we calculate how many moles of acid reacted, using the HBr solution's concentration and volume:

Molarity = Moles / Volume

Molarity * Volume = Moles

0.112 M * 12.4 mL = 1.389 mmol HBr

Now we convert HBr moles to Ca(OH)₂ moles, using the stoichiometric ratio:

1.389 mmol HBr * $\frac{1mmolCa(OH)_{2}}{2mmolHBr}$ = 0.6944 mmol Ca(OH)₂

Finally we calculate the molarity of the Ca(OH)₂ solution, using the given volume and calculated moles:

0.6944 mmol Ca(OH)₂ / 15.2 mL = 0.0457 M

7 0

3 years ago

The formation of ammonia is represented by the equation N2(g) + 3H2(g) ⇌ 2NH3(g). Determine the enthalpy of formation of ammonia

Savatey [412]

Answer:

$\Delta _fH_{NH_3}=-66\frac{kJ}{mol}$

Explanation:

Hello!

In this case, since the study of the bond energy allows us to compute the enthalpies of some reactions, for this combination reaction by which ammonia is yielded, we understand the enthalpy of reaction equals the enthalpy of formation of ammonia, and, in terms of the bonds energy we can write:

$\Delta _fH_{NH_3}=Delta _rH=\Sigma \Delta H(bonds \ broken)-\Sigma \Delta H(bonds \ formed)$

Whereas the bonds enthalpy of those bonds that get broken cover the N≡N and the three H-H bonds at the reactants side and the enthalpy of those bonds that are formed cover the six N-H bonds at the products; which means we obtain:

$\Delta _fH_{NH_3}=942\frac{kJ}{mol} +3*436\frac{kJ}{mol}-6*386\frac{kJ}{mol}\\\\\Delta _fH_{NH_3}=-66\frac{kJ}{mol}$

Which differs from the theoretical value that is -46 kJ/mol.

Best regards!

7 0

3 years ago