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3 years ago

How do you predict the geometrical shape of NH3 on VSEPR model

Chemistry

1 answer:

Sauron [17]3 years ago

6 0

Answer:

NH3 Ammonia

Explanation:

Ammonia has 4 regions of electron density around the central nitrogen atom (3 bonds and one lone pair). These are arranged in a tetrahedral shape. The resulting molecular shape is trigonal pyramidal with H-N-H angles of 106.7°.

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A sample of oxygen occupies 20.1 liters under a pressure of 1520 torr at 25.0o What volume would it occupy at 25.0oC if the pres

Zolol [24]

Answer:

The volume that the sample of oxygen would occupy at 25 ° C if the pressure were reduced to 760.0 torr is 40.2 L

Explanation:

Boyle's law establishes the relationship between the pressure and the volume of a gas when the temperature is constant, so that the pressure of a gas in a closed container is inversely proportional to the volume of the container. That is, if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

or P * V = k

Considering an initial state 1 and a final state 2, it is true:

P1* V1= P2*V2

In this case:

P1= 20.1 L
V1= 1520 torr
P2= 760 torr
V2= ?

Replacing:

20.1 L* 1520 torr= 760 torr* V2

Solving:

$V2=\frac{20.1 L* 1520 torr}{760 torr}$

V2= 40.2 L

The volume that the sample of oxygen would occupy at 25 ° C if the pressure were reduced to 760.0 torr is 40.2 L

4 0

2 years ago

During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of

almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate $\rm CaCO_3$ ?

Refer to a modern periodic table for relative atomic mass data:

Ca: 40.078;
C: 12.011;
O: 15.999.

Formula mass of $\rm CaCO_3$ :

$M(\mathrm{CaCO_3}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}$ .

$\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol$ .

How many moles of $\rm CaCl_2$ will be produced?

The coefficient in front of $\rm CaCO_3$ in the chemical equation is the same as that in front of $\rm CaCl_2$ . That is:

$\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1$ .

$\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol$ .

What's the theoretical yield of calcium chloride? In other words, what's the mass of $\rm 0.949184\;mol$ of $\rm CaCl_2$ ?

Again, refer to a periodic table for relative atomic data:

Ca: 40.078;
Cl: 35.45.

$M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}$ .

$\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}$ .

What's the actual yield of calcium chloride?

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%$ .

$\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}$ .

8 0

3 years ago

Other than reducing greenhouse gas emissions, why would an entity switch to an alternative energy solution

Alika [10]

Answer:

renewable energy sources such as solar and wind DONT emit carbon dioxide and other greenhouse gases that contribute to global warming

Explanation:

8 0

2 years ago

How many significant digits are in the measurement 64,000 mg A.2 B.3 C.4 D.5

Y_Kistochka [10]

5 because the 0's all matter.

8 0

3 years ago

A mixture of 5L of H2 and 3L O2 reacts to form H2O (g) at constant T and P .Find the

slamgirl [31]

The volume of H₂O = 5 L

<h3>Further explanation</h3>

Given

5L of H₂ and 3L O₂

Reaction

2H₂ (g) + O₂(g) ⇒2H₂O(g)

Required

The volume of H₂O

Solution

Avogadro's hypothesis:

In the same T,P and V, the gas contains the same number of molecules

So the ratio of gas volume will be equal to the ratio of gas moles

mol H₂ = 5, mol O₂ = 3

Find limiting reactants

From equation, mol ratio H₂ : O₂ = 2 : 1, so :

$\tt H_2\div O_2=\dfrac{mol~H_2}{coefficient}\div \dfrac{mol~O_2}{coefficient}\\\\=\dfrac{5}{2}\div \dfrac{3}{1}=2.5\div 3\rightarrow H_2~limiting~reactant(smaller~ratio)$

Find volume H₂O

mol H₂O based on mol H₂, and from equation mol ratio H₂ : H₂O=2 : 2, so mol H₂O = 5 mol and the volume also 5 L

4 0

2 years ago