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3 years ago

7

The brand names used on del monte, pillsbury, harley-davidson, and purina products are called _____ brands because of who owns t

hem

1 answer:

patriot [66]3 years ago

4 0

Manufacturers Is Your Answer

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1718 l of a 0.3556-m c3h7oh solution is diluted to a concentration of 0.1222 m, what is the volume of the resulting solution

Agata [3.3K]

Answer : The volume of the resulting solution will be 0.4999 L.

Explanation : As the two molar concentration of the $C_{3} H_{7}OH$ solution are given and one of the volume concentration needs to be found.

So, according to the formula :- $m_{1}V_{1} = m_{2}V_{2}$

And here if we consider $V_{1}$ as 0.1718 L moles and $m_{1}$ as 0.3556 moles then we need to find $V_{2}$ as $m_{2}$ is given as 0.1222 moles.
Hence on solving the formula we get,

$V_{2}$ = (0.1718 X 0.3556) / 0.1222 = 0.4999 L

5 0

4 years ago

A reddish layer of rust can form on old iron nails and bikes. In this chemical reaction, iron reacts slowly with oxygen in the a

Molodets [167]

Answer:

Iron slowly reacts with oxygen and forms rust. In this case, the reactants are iron and oxygen. The product is rust, or iron oxide. The chemical equation looks like this:

Explanation:

iron + oxygen → iron oxide

5 0

3 years ago

Read 2 more answers

While ethanol (CH3CH2OH) is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by

Anna [14]

<u>Answer:</u> The number of moles of ethanol after equilibrium is reached the second time is 11. moles.

<u>Explanation:</u>

We are given:

Initial moles of ethene = 34 moles

Initial moles of water vapor = 15 moles

The chemical equation for the formation of ethanol follows:

$CH_2=CH_2+H_2O\rightleftharpoons CH_3CH_2OH$

Initial: 34 15

At eqllm: 34-x 15-x x

We are given:

Equilibrium moles of ethene = 24 moles

Equilibrium moles of water vapor = 5 moles

Calculating for 'x'. we get:

$34-x=24\\\\x=10$

Volume of container = 100.0 L

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CH_3CH_2OH]}{[CH_2=CH_2][H_2O]}$ .......(1)

$[CH_3CH_2OH]=\frac{10}{100}=0.1M$

$[CH_2=CH_2]=\frac{24}{100}=0.24M$

$[H_2O]=\frac{5}{100}=0.05M$

Putting values in expression 1, we get:

$K_c=\frac{0.1}{0.24\times 0.05}\\\\K_c=8.3$

Now, 11 moles of ethene gas is again added and equilibrium is re-established, we get:

The chemical equation for the formation of ethanol follows:

$CH_2=CH_2+H_2O\rightleftharpoons CH_3CH_2OH$

Initial: 24+11 5 10

At eqllm: 35-x 5-x 10+x

$[CH_3CH_2OH]=\frac{10+x}{100}$

$[CH_2=CH_2]=\frac{35-x}{100}$

$[H_2O]=\frac{5-x}{100}$

Putting values of in expression 1, we get:

$8.3=\frac{\frac{(10+x)}{100}}{\frac{(35-x)}{100}\times \frac{(5-x)}{100}}\\\\8.3=\frac{(10+x)\times 100}{(35-x)\times (5-x)}\\\\x^2-52x+55=0\\\\x=50.9,1.1$

The value of 'x' cannot exceed '35', so the numerical value of x = 50.9 is neglected.

Moles of ethanol = $(10+x)=10+1.1=11.$

Hence, the number of moles of ethanol after equilibrium is reached the second time is 11. moles.

8 0

3 years ago

Scientist believe that brain and our sense of smell help us stay away from danger?

Vikki [24]

Yes because our instincts or senses tells us that its dangerous or something is going to happen

3 0

4 years ago

Read 2 more answers

Suppose now that you wanted to determine the density of a small crystal to confirm that it is sulfur. From the literature, you k

ivolga24 [154]

Answer:

Mix 11.73 mL of CHCl₃ and 8.27 mL of CHBr₃

Explanation:

To begin, it seems your question lacks the name of the pure samples that will be mixed to prepare the liquid mixture. The names are not necessary, as the numerical values are there. However, an internet search tells me they are CHCl₃ (d=1.492g/mL) and CHBr₃ (d=2.890 g/mL).

The mixture has a density of 2.07 g/mL, so <u>20 mL of the mixture would weigh</u>:

20 mL * 2.07 g/mL = 41.4 g

Let X be the volume of CHCl₃ and Y the volume of CHBr₃:

X + Y = 20 mL

The mass of CHCl₃ and CHBr₃ combined have to be equal to the mass of the mixture. We can write that equation using the volume of the samples and their density:

X * 1.492 + Y * 2.890 = 41.4 g

So now we have a system of two equations and two unknowns, we use algebra to solve it:

1. Express Y in terms of X:

X + Y = 20

Y = 20 - X

2. Replace Y in the second equation:

X * 1.492 + Y * 2.890 = 41.4

1.492*X + 2.890*(20-X) = 41.4

3. Solve for X:

1.492*X + 57.8 - 2.890*X = 41.4

1.398*X = 16.4

X = 11.73 mL

4. Using the now known value of X, solve for Y:

X + Y = 20

11.73 + Y = 20

Y = 8.27 mL

So, to prepare the liquid mixture we would mix 8.27 mL of CHBr₃ and 11.73 mL of CHCl₃.

7 0

4 years ago