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2 years ago

How many moles are there in 15.0 grams of SiO2

Chemistry

2 answers:

earnstyle [38]2 years ago

8 0

Answer:

0.25 M

Explanation:

There is 60.0843 G of SiO2 per M making 15 g equal roughly 0.25

Vinvika [58]2 years ago

7 0

n(mole)=mass: MM(molar mass)

n = 15 : 60

n = 0.25

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In the following reaction, how many grams of NaBr will produce 244 grams of NaNO3? Pb(NO3)2(aq) 2 NaBr(aq) PbBr2(s) 2 NaNO3(aq)

lyudmila [28]

Answer : The mass of $NaBr$ is, 295.323 grams

Solution :

First we have to calculate the moles of $NaNO_3$ .

$\text{Moles of }NaNO_3=\frac{\text{Mass of }NaNO_3}{\text{Molar mass of }NaNO_3}=\frac{244g}{85g/mole}=2.87moles$

Now we have to calculate the moles of NaBr.

The balanced chemical reaction is,

$Pb(NO_3)_2(aq)+2NaBr(aq)\rightarrow PbBr_2(s)+2NaNO_3(aq)$

From the balanced reaction we conclude that

As, 2 moles of $NaNO_3$ react with 2 moles of $NaBr$

So, 2.87 moles of $NaNO_3$ react with 2.87 moles of $NaBr$

Now we have to calculate the mass of NaBr.

$\text{Mass of }NaBr=\text{Moles of }NaBr\times \text{Molar mass of }NaBr$

$\text{Mass of }NaBr=(2.87mole)\times (102.9g/mole)=295.323g$

Therefore, the mass mass of $NaBr$ is, 295.323 grams

7 0

3 years ago

This exercise uses the radioactive decay model.

Anna11 [10]

For radioactive decay, we can relate current amount, initial amount, decay constant and time using:
N = No x exp(-λt)

Half-life = ln(2)/λ
λ = ln(2) / 5730
N/No = 80% = 0.8
0.8 = exp( -ln(2)/5730 x t)
t = 1844 years

3 0

3 years ago

Which compound reacts with an acid to produce water and a salt

Hatshy [7]

I'm not exactly sure which one but I do know that an acid and a base react in a aqueous solution to form water, so i would probably eliminate the ones that aren't aqueous solutions.

3 0

3 years ago

Read 2 more answers

For the generic equilibrium HA(aq) ⇌ H+(aq) + A−(aq), which of these statements is true? For the generic equilibrium , which of

timama [110]

<u>Answer:</u> The correct statement is if you add the soluble salt KA to a solution of HA that is at equilibrium, the pH would increase.

<u>Explanation:</u>

Common ion effect is defined as the effect which occurs on equilibrium when a common ion (an ion which is already present in the solution) is added to a solution. This effect generally decreases the solubility of a solute.

Equilibrium reaction of HA and KA follows the equation:

$HA\rightleftharpoons H^{+}(aq.)+A^{-}(aq.)$

$KA\rightleftharpoons K^+(aq.)+A^{-}(aq.)$

According to Le-Chateliers principle, if there is any change in the variables of the reaction, the equilibrium will shift in the direction in order to minimize the effect.

In the equilibrium reactions, $A^-$ ion is getting increased on the product side, so the equilibrium will shift in the direction to minimize this effect, which is in the direction of HA.

Thus, the addition of KA will shift the equilibrium in the left direction.

Equilibrium constant depends on the temperature of the system. It does not have any effect on any change of pH.

pH is defined as the negative logarithm of hydrogen ions present in the solution

If the solution has high hydrogen ion concentration, then the pH will be low.
If the solution has low hydrogen ion concentration, then the pH will be high.

As, the equilibrium is shifting in the left direction, that means concentration of $H^+$ ions are getting decreases. This will increase the pH of the solution.

Hence, the correct statement is if you add the soluble salt KA to a solution of HA that is at equilibrium, the pH would increase.

8 0

3 years ago

Is LiOH soluble or insoluble?

riadik2000 [5.3K]

LiOH is soluble. Na2CO3 is soluble. Cu(OH)2 is insoluble.

6 0

3 years ago