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2 years ago

HELP PLEASE I BEG OF YOU

Chemistry

1 answer:

Vitek1552 [10]2 years ago

7 0

Answer:

37.7%

Explanation:

58% of 65 is 37.7%

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Which of the following compounds has the highest boiling point?1. F22. NaF3. HF4. ClF

sergey [27]

Answer:

Explanation:

Hf has hydrogen bonding which is the strongest intermolecular forces. The stronger the IM forces, the higher the boiling point.

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2 years ago

What is the mole to mole relationship between copper and zinc? Cu2+(aq)+->Cu(s)+Zn2+(aq)

Aliun [14]

The reaction is missing the Zn(s) in the reactants. The stoichiometry of the copper/zinc is 1 mole to 1 mole

5 0

3 years ago

28: Which of the following alloys does not contain copper? 1.Duralumin 2. Bronze 3.Steel 4.Brass 5.Solder

Alja [10]

Answer:

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8 0

3 years ago

Sum of brother and sisters age is 26. Four times the brothers age is subtracted from three times the sisters age, the difference

lana66690 [7]

Answer : The ages of the brother and sister are, 10 and 16 years respectively.

Solution :

Let the age of brother be 'x' and the age of sister be 'y'

There are two equations formed which are,

x + y = 26 ..........(1)

3y - 4x = 8 ..........(2)

First we have to multiple equation (1) by 4 and then added equation (1) and (2), we get the value of 'y'.

$4x+4y+3y-4x=8+26\\\\7y=112\\\\y=16$

Now put the value of 'y' in equation (2), we get the value of 'x'

3(16) - 4x = 8

4x = 40

x = 10

The age of brother = x = 10 years

The age of sister = y = 16 years

Therefore, the ages of the brother and sister are, 10 and 16 years respectively.

3 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

2 years ago