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raketka [301]
3 years ago
12

Write formulas for the following compounds consisting of two non-metals or a metalloid and a non-metal.

Chemistry
1 answer:
Vladimir79 [104]3 years ago
3 0

Answer:

Explanation:

1) HCl

2) H₂S

3) HI

4) SiF₄

5 ) AsF₅

6) XeF₆

7 ) IF₅

8 ) KrF₂

9 ) S₄N₄

10) Cl₂O₇

11) PH₃

12) P₄O₁₀

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How fast do i have to run 250 m in 30 seconds
elena-14-01-66 [18.8K]

Answer:

S=D/T

250/30= 8.33 miles

Explanation:

7 0
3 years ago
if a baloon filled with dry hydrogen weighs 35 gram,but weighs 440 grams when filled with the vapour of an organic compound. cal
Elan Coil [88]

Answer:

1) The vapor density of the organic compound is approximately 12.57

2) The relative molar mass (RMM) of the organic compound is approximately 25.14 grams  

Explanation:

1) The mass of the balloon filled with dry hydrogen = 35 grams

The mass of the balloon filled with vapor of an organic compound = 440 grams

The vapor density = (Weight of a given volume of gas)/(Weight of equal volume of hydrogen)

The vapor density of the organic compound = (440)/(35) ≈ 12.57

The vapor density of the organic compound ≈ 12.57

2) The relative molar mass (RMM) = 2 × vapor density

The relative molar mass (RMM) of the organic compound = 2 × vapor density of the organic compound

The relative molar mass (RMM) of the organic compound ≈ 2 × 12.57 ≈ 25.14 grams  

The relative molar mass (RMM) of the organic compound ≈ 25.14 grams  

6 0
3 years ago
The chart shows the solubility of different substances.
Annette [7]
Answer: 1) Temperature can change the solubility of a solute.

Explanation:

The chart is missing so there is no way to tell what does the graph show.

Yet, I can help you because I can explain the status of each statement of the choices. As you will see there is only one possibility..

<span>1) Temperature can change the solubility of a solute.

Yes, temperature definetly can, and mostly do, modify the solubility of a solute.

You can search any chart of solubility and will find that.

I can give you two examples:

a) Sodium chloride: dissolve some spoons of salt in a cold water  until you can not dissolve more. Then, heat the water, you will find that more salt will get dissolved, proving that the temperature of the solution increases the solubility of sodium chloride.

b) Carbon dioxide gas: the soft drinks have CO₂ molecules dissolved in it.
 
The higher the temperature of the soft drink the less the amount of CO₂(g) that can be dissolved. That is why the soda bottling plants cool the beverage before adding the CO₂(g).

2) </span><span>Temperature has no affect on the solubility of a solute.

Since this is the opposite to the first statement and the first is true, this is false.

3) Salt has a greater solubility than sugar.

False.

This is an empirical result, which you cannot predict theoretically. So you need to see at the data either in a table or in a chart. Else you can test it at home. After the empirical data are shown it results that more grams of sugar can be dissolved in water compared to salt.

That is something you ca see in a chart or you can prove by yourself.

4) Nitrite salt has a greater solubility than sugar. </span>

False.

Looking at some data you can find that sodium nitrite solutiliby is aroun  70 - 100 g/10 g while sugar (sucrose) solutiblity is around 180 - 235 g/ 100 g.

8 0
4 years ago
Read 2 more answers
4.60 mL of 0.1852 M HNO3 is titrated to the phenolphthalein indicator endpoint with 27.35 mL of a KOH solution. What is the mola
Kamila [148]

Answer:

M_{base}=0.0311M

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve for the molarity of the KOH solution by knowing this base react in a 1:1 mole ratio with nitric acid, HNO3; thus, we can write the following equation, as their moles are the same at the endpoint:

n_{acid}=n_{base}

Which in terms of molarities and volumes is:

M_{acid}V_{acid}=M_{base}V_{base}

Thus, we solve for the molarity of the base (KOH) to obtain:

M_{base}=\frac{M_{acid}V_{acid}}{V_{base}} \\\\M_{base}=\frac{4.60mL*0.1852M}{27.35mL}\\\\M_{base}=0.0311M

Regards!

4 0
3 years ago
For the hypothetical reaction: A + B ---&gt; 2C, write an expression that relates the disappearance of A and B to the appearance
forsale [732]

Answer:

Rate expression has been given below

Explanation:

According to the given equation, 1 molecule of A reacts with 1 molecule of B and produces 2 molecules of B at a time.

So, rate of disappearance of both A and B are one half of rate of appearance of B

Hence rate expression can be represented as:

Rate=\frac{-\Delta [A]}{\Delta t}=\frac{-\Delta [B]}{\Delta t}=\frac{1}{2}\frac{\Delta [C]}{\Delta t}

where \frac{-\Delta [A]}{\Delta t} is rate of disappearance of A, \frac{-\Delta [B]}{\Delta t} is rate of disappearance of B and \frac{\Delta [C]}{\Delta t} rate of appearance of C

8 0
4 years ago
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