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3 years ago

Ammonium hydroxide and sulfuric acid react to produce?

Chemistry

1 answer:

krek1111 [17]3 years ago

8 0

Answer:

Explanation:

The reactions of acids and bases always produce heat and a salt

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How many molecules of ethane are present in 64.28 liters of ethane gas (C2H6) at STP

Rom4ik [11]

Answer : The molecule of ethane present in 64.28 L of ethane gas at STP is, $17.277\times 10^{23}$ molecule.

Solution :

At STP,

22.4 L volume of ethane present in 1 mole of ethane gas

64.28 L volume of ethane present in $\frac{64.28L}{22.4L}\times 1mole=2.869moles$ of ethane gas

And, as we know that

1 mole of ethane molecule contains $6.022\times 10^{23}$ molecules of ethane

2.869 moles of ethane molecule contains $2.869\times 6.022\times 10^{23}=17.277\times 10^{23}$ molecules of ethane

Therefore, the molecule of ethane present in 64.28 L of ethane gas at STP is, $17.277\times 10^{23}$ molecule.

4 0

3 years ago

Read 2 more answers

How many grams of NaoH are needed to nutralize 50 grams of H2SO4?

sergij07 [2.7K]

to neutralize 1 mole of H2 S o4 we need one mole of any if we are having 50 grams of H2 S o4 it means the mole of H2 S o4 in 50 gram will be 50×40)/98 hence

utilising 50 grams of H2 S o4 we need approximately 20. 5 gm of Naoh

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3 years ago

What volume in mL of 0.220 M HBr solution is required to produce 0.0130 moles of HBr?

yulyashka [42]

Answer: A volume of 59 mL of 0.220 M HBr solution is required to produce 0.0130 moles of HBr.

Explanation:

Given: Moles = 0.0130 mol

Molarity = 0.220 M

Molarity is the number of moles of solute present in liter of a solution.

$Molarity = \frac{moles}{volume (in L)}$

Substitute the values into above formula as follows.

$Molarity = \frac{moles}{volume (in L)}\\0.220 M = \frac{0.0130 mol}{Volume (in L)}\\Volume (in L) = 0.059 L$

As 1 L = 1000 mL

So, 0.059 L = 59 mL

Thus, we can conclude that a volume of 59 mL of 0.220 M HBr solution is required to produce 0.0130 moles of HBr.

8 0

3 years ago

What experimental evidence did thomson use to support his hypothesis that atoms of all elements are made up of even smaller, cha

Serggg [28]

<span>Identical particles were produced no matter which metal he used.</span>

4 0

3 years ago

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In the laboratory you dissolve 19.1 g of ammonium fluoride in a volumetric flask and add water to a total volume of 375 . mL. Wh

11111nata11111 [884]

Answer:

1.376 M

Explanation:

The following data were obtained from the question:

Mass of ammonium fluoride (NH₄F) = 19.1 g

Volume of solution = 375 mL

Molarity of ammonium fluoride (NH₄F) =?

Next, we shall convert 375 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

375 mL = 375 mL × 1 L / 1000 mL

375 mL = 0.375 L

Next, we shall determine the number of mole in 19.1 g of ammonium fluoride (NH₄F). This can be obtained as follow:

Mass of NH₄F = 19.1 g

Molar mass of NH₄F = 14 + (4×1) + 19

= 14 + 4 + 19

Molar mass of NH₄F = 37 g/mol

Mole of NH₄F =?

Mole = mass /Molar mass

Mole of NH₄F = 19.1 / 37

Mole of NH₄F = 0.516 mole

Finally, we shall determine the molarity of the solution. This can be obtained as follow:

Volume of solution = 0.375 L

Mole of NH₄F = 0.516 mole

Molarity of NH₄F =?

Molarity = mole /Volume

Molarity of NH₄F = 0.516 / 0.375

Molarity of NH₄F = 1.376 M

Therefore, the molarity of the ammonium fluoride (NH₄F) is 1.376 M

3 0

2 years ago