Which one is an ionic compound? A. CINe B. NaP C. Na2 D. NO E. FeW

For each example, identify if the food contains mostly saturated or unsaturated fatty acids.

dedylja [7]

Answer:

1. Vegetable oil, which is a liquid at room temperature, is used in making cake batter. Unsaturated

2. Chocolate, which is solid at room temperature, is derived from a seed. Saturated

3. Butter, which is a solid at room temperature, is often used in cooking. Saturated

4. The oil inside flaxseed oil pills is liquid at room temperature. Unsaturated

Explanation:

Saturated fats are those kinds of fat that do not have a double bond between their carbon atoms when represented structurally. At room temperature, they tend to assume the solid-state, and they are mostly obtained from animal sources. Unsaturated fats do have a minimum of one double bond between their carbon atoms and tend to assume the liquid phase. They are mostly obtained from plant sources.

1. Vegetable oil is obtained from a plant source and is, therefore, unsaturated fat.

2. Chocolate even though it comes from cocoa is solid at room temperature and is mostly comprised of saturated fat.

3. Butter a solid at room temperature is mostly made up of saturated fat and is sourced from animals.

4. Flaxseed oil obtained from a plant source is mostly unsaturated fat.

3 0

3 years ago

A student adds 200.0g of C7H6O3 to an excess of C4H6O3, this produces C9H8O4 and C2H4O2. Calculate the percent yield if 231 g of

kari74 [83]

Answer: Percent yield = 88%

Explanation:

Given data:

Mass of C₇H₆O₃ = 200.0 g

Actual yield of aspirin = 231 g

Percent yield = ?

Solution:

Chemical equation:

C₇H₆O₃ + C₄H₆O₃ → C₉H₈O₄ + C₂H₄O₂

Number of moles of aspirin:

Number of moles = mass/molar mass

Number of moles = 200.0 g/ 138.12 g/mol

Number of moles = 1.45 mol

Now we will compare the moles of aspirin with C₇H₆O₃.

C₇H₆O₃ : C₉H₈O₄

1 : 1

1.45 : 1.45

Theoretical yield of aspirin:

Mass = number of moles × molar mass

Mass = 1.45 g × 180.158 g/mol

Mass = 261.23 g

Percent yield:

Percent yield =( actual yield / theoretical yield )× 100

Percent yield = (231 g/ 261.23 g)× 100

Percent yield = 0.88 × 100

Percent yield = 88%

:

7 0

2 years ago

What is the frequency of a radio wave with a wavelength of 578.0m

Radda [10]

Answer:

0.0017 Hz

Explanation:

the solution is in the photo

4 0

3 years ago

Determine the resulting pH when 12mL if 0.16M HCl are reacted with 32 mL if 0.24M KOH.

TEA [102]

Answer:

pH = 13.1

Explanation:

Hello there!

In this case, according to the given information, we can set up the following equation:

$HCl+KOH\rightarrow KCl+H_2O$

Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:

$n_{HCl}=0.012L*0.16mol/L=0.00192mol\\\\n_{KOH}=0.032L*0.24mol/L=0.00768mol$

Thus, since there are less moles of HCl, we calculate the remaining moles of KOH as follows:

$n_{KOH}=0.00768mol-0.00192mol=0.00576mol$

And the resulting concentration of KOH and OH ions as this is a strong base:

$[KOH]=[OH^-]=\frac{0.00576mol}{0.012L+0.032L}=0.131M$

And the resulting pH is:

$pH=14+log(0.131)\\\\pH=13.1$

Regards!

3 0

3 years ago

A chemist titrates 120.0 mL of a 0.4006 M hydrocyanic acid (HCN) solution with 0.6812 MNaOH solution at 25 °C. Calculate the pH

lukranit [14]

Explanation:

The given reaction equation is as follows.

$HCN + NaOH \rightarrow NaCN + H_{2}O$

Hence, initial moles of HCN will be as follows.

Moles = Molarity × Volume

= $0.4006 M \times 120 ml$

= 48.072 mol

Now, we will calculate the volume of NaOH as follows.

$M_{1}V_{1} = M_{2}V_{2}$

$0.4006 \times 120 = 0.6812 \times V_{2}$

$V_{2}$ = 70.57 ml

At the equivalence point, moles of both HCN and NaOH will be equal. And, total volume will be as follows.

120 ml + 70.57 ml = 190.57 ml

Initial concentration of $CN^{-}$ is as follows.

$\frac{48.072}{190.57}$

= 0.25 M

Now, the equilibrium equation will be as follows.

$CN^{-} + H_{2}O \rightleftharpoons HCN + OH^{-}$

Initial: 0.25

Equilbm: 0.25 - x x x

Expression to find $K_{a}$ is as follows.

$K_{a} = \frac{x^{2}}{0.25 - x}$

We know that, $pK_{a} = -log K_{a}$

$K_{a}$ = antilog (-9.21)

= $6.16 \times 10^{-10}$

As, $K_{b} = \frac{K_{w}}{K_{a}}$

$\frac{10^{-14}}{6.16 \times 10^{-10}} = \frac{x^{2}}{0.25 - x}$

x = $0.2 \times 10^{-2} = [OH^{-}]$

Now, we will calculate the pH as follows.

pOH = $-log[OH^{-}]$

as pOH = 2.698

And, pH + pOH = 14

pH = 14 - 2.698

= 11.30

Thus, we can conclude that pH at equivalence is 11.30.

4 0

3 years ago