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Arlecino [84]
3 years ago
9

Help me pleaseeeeeeeeeeeeeeee

Chemistry
1 answer:
Burka [1]3 years ago
8 0

Answer:

C) 2-bromo-1-methylpropane

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Suppose now that you wanted to determine the density of a small crystal to confirm that it is phosphorus. From the literature, y
denis23 [38]

Answer:

Volume of CHCl_3   = 15.31 mL

Volume of CHBr_3 = 4.69 mL

Explanation:

Given that:

the density of the mixture = 1.82 g/mL

From the density of the pure samples

The density of CHCl_3 = 1.492 g/mL

The density of CHBr_3 = 2.890 g/mL

The total volume of the liquid mixture = 20.0 mL

Suppose the volume of  CHCl_3 = P ml

and the volume of CHBr_3 = Q ml

the sum of their volumes should be equal to the total volume of the mixture

P \ ml + Q \ ml = 20 ml  ----- (1)

However, we know that Density = mass/volume

∴ mass = density × volume

The equation can now be expressed as:

\mathtt{(Density \ of  \ CHCl_3 \times Vol. \ of  \ CHCl_3 ) + (Density  \  of \  CHBr_3  \times \ volume \ of \ CHBr_3)} = \mathtt{ (Density  \ of \ mixture \times volume \ of \ the \ mixture)}

1.492 g/mL × P mL + 2.890 g/mL × Q mL = 1.82 g/mL × 20 mL  ---- (2)

From equation (1) ;

let Q = 20 - P

The replace the value of P into equation (2)

1.492 g/mL × P mL + 2.890g/mL × (20 - P) mL = 1.82 g/mL × 20 mL

1.492 P g + 57.8g - 2.890 P g =  36.4g

1.492 P g - 2.890 P g = 36.4g - 57.8g

-1.398 P g = -21.4g

P = -21.4g/-1.398g

P = 15.31 mL

Q = 20 - P

Q = (20 - 15.31) mL

Q = 4.69 mL

∴

Volume of CHCl_3   = 15.31 mL

Volume of CHBr_3 = 4.69 mL

6 0
3 years ago
Which parts of sedimentary rock formation include the breakdown and carrying away of existing rock? Check all that apply.
stiks02 [169]

Answer

Weathering and erosion are two processes of which sedimentary rocks are broken down and carry away existing rocks.

Explanation:

3 0
3 years ago
A solution of starch at room temperature does not readily decompose to form a solution of simple sugars because
Basile [38]

Explanation:

The starch requires a temperature higher than the room temperature (arround 60 °C) to decompose to form simple sugars. This is because the energy required to break the chemical bonds. Also, it may need the action of some specific enzymes (alpha and beta amilase) to break those bonds.

6 0
3 years ago
What atomic component contributes to many of the trends of elements
aivan3 [116]

what??? i need more information

3 0
3 years ago
PCl3(g) + Cl2(g) ⇋ PCl5(g) Kc = 91.0 at 400 K. What is the [Cl2] at equilibrium if the initial concentrations were 0.24 M for PC
Dmitry_Shevchenko [17]

Answer:

[Cl₂] in equilibrium is 1.26 M

Explanation:

This is the equilibrium:

PCl₃(g) + Cl₂(g) ⇋ PCl₅(g)

Kc = 91

So let's analyse, all the process:

                PCl₃(g)        +        Cl₂(g)     ⇋        PCl₅(g)

Initially     0.24 M                 1.50M                 0.12 M

React           x                           x                         x

Some amount of compound has reacted during the process.

In equilibrium we have

              0.24 - x                  1.50 - x                  0.12 + x

As initially we have moles of product, in equilibrium we have to sum them.

Let's make the expression for Kc

Kc = [PCl₅] / [Cl₂] . [PCl₃]

91 = (0.12 + x) / (0.24 - x) ( 1.50 - x)

91 = (0.12 + x) / (0.36 - 0.24x - 1.5x + x²)          

91 (0.36 - 0.24x - 1.5x + x²) = (0.12 + x)

32.76 - 158.34x + 91x² = 0.12 +x

32.64 - 159.34x + 91x² = 0

This a quadratic function:

a = 91; b= -159.34; c = 32.64

(-b +- √(b² - 4ac)) / 2a

Solution 1 = 1.5

Solution 2 = 0.23 (This is our value)

So [Cl₂] in equilibrium is 1.50 - 0.23 = 1.26 M

5 0
3 years ago
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