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3 years ago

What is the x-coordinate of the function below when the y-coordinate is 0?(0,5) (2,-5)

Chemistry

1 answer:

maria [59]3 years ago

4 0

The equation of the function is assumed to be a straight line is of the form,

$y= mx+c$ , where $m,c$

are the gradient and the y intercept respectively.

The gradient is computed using the formula for the gradient, i.e

$m=\frac{y_2-y_1}{x_2-x_1} = \frac{-5-5}{2-0} = -5$ .

The equation of the line is,

$y= -5x+c\\=>5=-5(0)+c\\=> c= 5$

The equation of the function is ,

$y = -5x+5.$ .

The value of the x-coordinate when the y coordinate is 0 is found as shown below.

$0=-5x+5\\=>x=1\\$ .

so the x-coordinate is 1.

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A 20.0-milliliter sample of 0.200 M K2CO3 solution is added to 30.0 milliliters of 0.400 M Ba(NO3)2 solution.

jenyasd209 [6]

Answer:

(B) 0.160 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $K_2CO_3$ :

Molarity = 0.200 M

Volume = 20.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 20.0×10⁻³ L

Thus, moles of $K_2CO_3$ :

$Moles=0.200 \times {20.0\times 10^{-3}}\ moles$

Moles of $K_2CO_3$ = 0.004 moles

For $Ba(NO_3)_2$ :

Molarity = 0.400 M

Volume = 30.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume =30.0×10⁻³ L

Thus, moles of $Ba(NO_3)_2$ :

$Moles=0.400\times {30.0\times 10^{-3}}\ moles$

Moles of $Ba(NO_3)_2$ = 0.012 moles

According to the given reaction:

$K_2CO_3_{(aq)}+Ba(NO_3)_2_{(aq)}\rightarrow BaCO_3_{(s)}+2KNO_3_{(aq)}$

1 mole of potassium carbonate react with 1 mole of barium nitrate

0.004 moles potassium carbonate react with 0.004 mole of barium nitrate

Moles of barium nitrate = 0.004 moles

Available moles of barium nitrate = 0.012 moles

Limiting reagent is the one which is present in small amount. Thus, potassium carbonate is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of potassium carbonate gives 1 mole of barium carbonate

Also,

0.004 mole of potassium carbonate gives 0.004 mole of barium carbonate

Mole of barium carbonate = 0.004 moles

Also, consumed barium nitrate = 0.004 moles (barium ions precipitate with carbonate ions)

Left over moles = 0.012 - 0.004 moles = 0.008 moles

Total volume = 20.0 + 30.0 mL = 50.0 mL = 0.05 L

So, Concentration = 0.008/0.05 M = 0.160 M

(B) is correct.

4 0

3 years ago

A cylinder and piston assembly (defined as the system) is warmed by an external flame. The contents of the cylinder expand, doin

melomori [17]

Answer:

ΔE = 73 J

Explanation:

By the first law of thermodynamics, the energy in the system must conserved:

ΔE = Q - W

Where ΔE is the internal energy, Q is the heat flow (positive if it's absorbed by the system, and negative if the system loses heat), and W is the work (positive if the system is expanding, and negative if the system is compressing).

So, Q = + 551 J, and W = + 478 J

ΔE = 551 - 478

ΔE = 73 J

3 0

3 years ago

Ethanol, C2H6O, is most often blended with gasoline - usually as a 10 percent mix - to create a fuel called gasohol. Ethanol is

weeeeeb [17]

Answer:

This means 463 grams of ethanol would provide less amount of energy

Explanation:

Step 1: Data given

Heat of combustion of ethanol = 326.7 kcal/mol

The heat of combustion of octane = 1.308*10³ kcal/mol

Mass of octane = 463 grams

Molar mass octane = 114.23 g/mol

Molar mass ethanol = 46.07 g/mol

Step 2: Calculate moles octane

Moles octane = mass octane / molar mass octane

Moles octane = 463 grams / 114.23 g/mol

Moles octane = 4.05 moles

Step 3: Calculate energy of combustion of 4.05 moles octane

Combustion of 1 mol octane gives us: 1.308 * 10³ kcal/mol

Combustion of 4.05 moles octane gives us 4.05 * 1.308 * 10³ kcal/mol = 5.30 * 10³ kcal

This means the combustion reaction of 463 grams of octane gives us 5.30 * 10³ kcal

Step 4:

Heat of combustion of ethanol = 326.7 kcal/mol

OR in words: combustion of 1 mol ethanol gives us 326.7 kcal energy

Moles ethanol = 463 grams / 46.07 g/mol

Moles ethanol = 10.05 moles

Since combustion of 1 mol ethanol gives us 326.7 kcal

10.05 moles ethanol will give us = 10.05 * 326.7 = 3283.3 kcal = 3.28 * 10³ kcal

5.30 * 10³ kcal > 3.28 * 10³ kcal

This means 463 grams of ethanol would provide less amount of energy

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3 years ago

The freezing point of a substance is -20°C. Its boiling point is 120°C.

Tasya [4]

Answer:

a. liquid

b. solid

c. gas, (should be at it's boiling point)

Explanation: If the normal melting point of a substance is below room temperature, the substance is a liquid at room temperature. Benzene melts at 6°C and boils at 80°C; it is a liquid at room temperature. If both the normal melting point and the normal boiling point are above room temperature, the substance is a solid.

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3 years ago

Reactions cannot occur without a certain minimum amount of energy. What is this minimum amount of energy called? A.the activatio

madam [21]

A. the activation energy

7 0

3 years ago

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