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2 years ago

NEED ANSWER ASAP! How many grams are in 1.50 moles of bromine liquid?

Chemistry

1 answer:

padilas [110]2 years ago

7 0

Answer:

119.85 grams Br or 120. grams Br (sig figs)

Explanation:

1.50 moles Br 79.90 g Br

--------------------- x ------------------------ = 119.85 grams Br or 120 grams Br (sig figs)

1 mole

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Which scientist discovered the neutron and gave us the current basic model of

Evgesh-ka [11]

Answer:

Explanation:

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3 years ago

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lisabon 2012 [21]

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2 years ago

If you wanted to protect water from freezing, which of the compounds below would accomplish this best. NaNO3, Mg(NO3)2, or Al(NO

Korvikt [17]

All of them are soluble salt.

First one dissociates into two ions.

The second one dissociates into 3 ions.

The third dissociate into 4 ions. therefore, Al(NO3)3

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2 years ago

The density of water at 400C is 0.992 g/mL What is the volume of 27.0 g of water at this temperature?

pantera1 [17]

Answer:

Volume of water at this temperature is 27.2 mL

Explanation:

We know that $density=\frac{mass}{volume}$

Here density of water is 0.992 g/mL

Here mass of water is 27.0 g

So $volume=\frac{mass}{density}$

= $\frac{27.0g}{0.992g/mL}$

= 27.2 mL

7 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

2 years ago