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gayaneshka [121]
4 years ago
8

What happens to a line when the y-intercept is changed? Check all that apply.

Chemistry
1 answer:
tamaranim1 [39]4 years ago
5 0

Answer:

As the y-intercept increases, the graph of the line shifts up;

As the y-intercept decreases, the graph of the line shifts down

Explanation:

There are two ways to think about this problem. The first way would be the graphical approach:

  • if we only change the y-intercept, this means we keep the same slope;
  • y-axis is the vertical axis;
  • if we change the point at which the line crosses the y-axis, we either shift it upward for a higher y-intercept or downward for a lower y-intercept.

Now, thinking algebraically, a line has the following equation in a general form:

y = mx + b

The y-intercept is essentially obtained when x = 0, then:

y = b:

  • if we increase b value, the y value increases, so the graph shifts upward;
  • if we decrease b value, the y value decreases, so the graph shifts downward.
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A 5.5 g sample of a substance contains only carbon and oxygen. Carbon makes up 35% of the mass of the substance. The rest is mad
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We have been given the condition that carbon makes up 35% of the mass of the substance and the rest is made up of oxygen. With this, it can be concluded that 65% of the substance is made up of oxygen. If we let x be the mass of oxygen in the substance, the operation that would best represent the scenario is,

<span>                                       x = (0.65)(5.5 g)</span>

<span>                                       <em> </em><span><em>x = 3.575 g</em></span></span>

8 0
3 years ago
Formula for the compound that contains Mg2+ and O2-
jeka94

Answer:

MgO.

Explanation:

charges of both satisfy one another (balanced) -- producing a compound MgO.

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3 years ago
Predict whether ΔS° is greater than, less than, or approximately zero for each of the following reactions, and explain your choi
inna [77]

Answer:

Explanation:

Entropy -

In a system, the randomness is measured by the term entropy .

Randomness basically refers as a form of energy that can not be used for any work.

The change in entropy is given by amount heat per change in temperature.

  • When solid is converted to gas entropy increases,

As the molecules in solid state are tightly packed and has more force of attraction between the molecules, but as it is converted to gas, the force of attraction between the molecule decreases and hence entropy increases.

So,

The particles of the substance , if are tightly held by strong force of attraction will decrease the entropy ,

And

If the particles are loosely held , the entropy will increase .  

  • If in a reaction , more number of gaseous atoms are present in the product side , entropy will increase , i.e. Δ°S > 0
  • When liquid is converted to solid entropy decreases,  

As the molecules in liquid state are loosely packed and has less force of attraction between the molecules, but as it is converted to solid, the force of attraction between the molecule increases and hence entropy decreases.

  • If in a reaction , less number of gaseous atoms are present in the product side , entropy will decrease , i.e. Δ°S < 0

From the question ,

( a )  NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)

Gaseous atoms -

Reactant - 1 + 5 = 6

Product - 4 + 6 = 10 ,

Hence ,

More number of gaseous atoms are present in the product side , So ,

entropy will increase , i.e. Δ°S > 0

( b ) CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g)

Gaseous atoms -

Reactant - 1 + 2 = 3

Product - 1 + 2 = 3 ,

Since ,

Both the side the value of gaseous atoms are , hence , Δ°S = 0 .

( c ) CaCO₃(s) → CaO(s) + CO₂(g)

Gaseous atoms -

Reactant = 0

Product - 0 + 1 = 1 ,

Since ,

Hence ,

More number of gaseous atoms are present in the product side , So ,

entropy will increase , i.e. Δ°S > 0

8 0
4 years ago
Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and
skelet666 [1.2K]

Answer : The values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 33.89kJ,95.94J/K\text{ and }5.299kJ/mol respectively.

Explanation :

The given balanced chemical reaction is,

C_6H_6(l)\rightarrow C_6H_6(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0_{(C_6H_6(g))} = standard enthalpy of formation  of gaseous benzene = 82.93 kJ/mol

\Delta H_f^0_{(C_6H_6(l))} = standard enthalpy of formation  of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]

\Delta H^o=33.89kJ/mol=33890J/mol

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S^0_{(C_6H_6(g))} = standard entropy of formation  of gaseous benzene = 269.2 J/K.mol

\Delta S^0_{(C_6H_6(l))} = standard entropy of formation  of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]

\Delta S^o=95.94J/K.mol

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 25^oC\text{ or }298K.

\Delta G^o=(33890J)-(298K\times 95.94J/K)

\Delta G^o=5299.88J/mol=5.299kJ/mol

Therefore, the values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 33.89kJ,95.94J/K\text{ and }5.299kJ/mol respectively.

6 0
3 years ago
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Mnenie [13.5K]

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Explanation:

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2 years ago
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