Please help me asap!

calculate how many Liters of 0.50 M silver nitrate solution you will need to provide the 2.4x10^-3 moles of silver nitrate

wlad13 [49]

Answer:

4.8x10⁻³ Liters are required

Explanation:

Molarity is an unit of concentration in chemistry defined as the ratio between moles of solute (In this case, silver nitrate) and liters of solution.

The 0.50M solution contains 0.50 moles of silver nitrate per liter of solution.

To provide 2.4x10⁻³ moles Silver nitrate are required:

2.4x10⁻³ moles * (1L / 0.50 moles) =

<h3>4.8x10⁻³ Liters are required</h3>

8 0

3 years ago

32. In general, ionization energy increases from left to right across a given

miskamm [114]

Aluminum's outer electron in 3p-orbital is slightly further from the nucleus and protected by 3s-orbital. This causes it to need less energy. Magnesium's outer electron is closer and not protected by an outer orbital.
Hope This Helps and Ged Bless!

6 0

3 years ago

Which state of matter takes both the shape and volume of its container?

NISA [10]

10. is C or gas and 11. is B liquid.

8 0

3 years ago

Read 2 more answers

what is the percent yield of titanium (II) oxide if 20.0 grams of titanium (II) sulfide is reacted with water? The actual yield

earnstyle [38]

Answer : The percent yield of titanium (II) oxide is, 142.5 % and the impurities could have caused the percent yield to be so high.

Explanation : Given,

Mass of titanium(II) sulfide = 20.0 g

Molar mass of titanium(II) sulfide = 79.9 g/mole

Molar mass of titanium(II) oxide = 63.9 g/mole

First we have to calculate the moles of titanium(II) sulfide.

$\text{ Moles of titanium(II) sulfide}=\frac{\text{ Mass of titanium(II) sulfide}}{\text{ Molar mass of titanium(II) sulfide}}=\frac{20.0g}{79.9g/mole}=0.2503moles$

Now we have to calculate the moles of titanium(II) oxide.

The balanced chemical reaction is,

$TiS+H_2O\rightarrow TiO+H_2S$

From the reaction, we conclude that

As, 1 mole of titanium(II) sulfide react to give 1 mole of titanium(II) oxide

So, 0.2503 mole of titanium(II) sulfide react to give 0.2503 mole of titanium(II) oxide

Now we have to calculate the mass of titanium(II) oxide.

$\text{ Mass of titanium(II) oxide}=\text{ Moles of titanium(II) oxide}\times \text{ Molar mass of titanium(II) oxide}$

$\text{ Mass of titanium(II) oxide}=(0.2503moles)\times (63.9g/mole)=15.99g$

To calculate the percentage yield of titanium (II) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of titanium (II) oxide = 22.8 g

Theoretical yield of titanium (II) oxide = 15.99 g

Putting values in above equation, we get:

$\%\text{ yield of titanium (II) oxide}=\frac{22.8g}{15.99g}\times 100\\\\\% \text{yield of titanium (II) oxide}=142.5\%$

Hence, the percent yield of titanium (II) oxide is, 142.5 %

If the percent yields is greater than 100% that means the product of the reaction contains impurities which cause its mass to be greater than it actually.

5 0

3 years ago

2. The power steering in an automobile has a mechanical advantage of roughly 75. If the input force on the steering wheel is 49

sertanlavr [38]

1.53

Explanation:

49/x=75 X or output = 1.53

5 0

3 years ago