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3 years ago

What is the new pressure if a gas takes up 350.0 mL of space at 745.0 mmHg and is pushed into 340.0 mL of space?

Chemistry

1 answer:

dolphi86 [110]3 years ago

4 0

Hope this helps you please mark as brainliest

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Which letter corresponds with the location of f orbitals on the periodic table?

Lerok [7]

Answer:

The correct would be C i think :)

Explanation:

Stay postivie :)

8 0

3 years ago

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa

notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

$2NaOH+H_2SO_4-->Na_2SO_4+2H_2O$

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

$n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\$

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

$0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4$

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

$0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4$

Finally, the percent yield turns out into:

$Y=\frac{1.92g}{7.1g} *100$

$Y=27.0$ %

Best regards.

6 0

3 years ago

Unknown # 41

sveta [45]

Answer:

Explanation:

From the question, we can see that a qualitative analysis of the compound shows that it has a lilac flame colour. The lilac flame colour corresponds to the potassium ion (K^+).

Again, the test of addition of HNO3(aq) and AgNO3(aq) to a solution is a test for halogens. If the result is a green precipitate, then the ion present is the iodide ion (I^-).

Hence, the compound must be KI.

4 0

3 years ago

Help me with this please

MrRissso [65]

Answer:

I think B no is correct orterwise Ano

8 0

3 years ago

If a chemist wants to make 1.3 L of 0.25 M solution of KOH by diluting a stock solution of 0.675 M KOH, how many milliliters of

anzhelika [568]

To solve this we use the equation,

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

.675 M x V1 = .25 M x 1.3 L

V1 = 0.48 L or 480 mL

8 0

3 years ago