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Sunny_sXe [5.5K]
3 years ago
13

2. Fill in the blanks with an example of each type of matter,

Chemistry
1 answer:
Korvikt [17]3 years ago
6 0

Explanation:

Matter- gas

Substance-o2

Mixture-sugar and water

Compound-BeCl2

Element-Mg

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In the late 1400s,ties among people of the same formed the basics of most aspects of life in rural west Africa
lara31 [8.8K]
Can u write it more specifically? so that I can answer
6 0
3 years ago
The following reaction has an equilibrium constant (Keq) of 1 under standard conditions and can be catalysed by the enzyme aspar
Trava [24]

Answer:

  1. What is the AGⓇ of this reaction? 0.
  2. Which will be favoured - the forward reaction, the reverse reaction, or neither? Neither.
  3. What effect does the presence of the enzyme aspartate transaminase have on the Key value when compared with its value in the absence of enzyme? It does not affect the value of Keq.
  4. If one of the products of reaction 1, oxaloacetate, is removed by converting it to citrate as follows: Reaction 2: oxaloacetate + acetyl-CoA citrate + COASH will the key for Reaction l be changed? No, the Keq does not change.

Explanation:

1. To calculate the delta G of a reaction given the K, we use the following equation:

ΔG°= -RT ln K.

Which gives us 0 when K is 1.

2.None of the reactions is favoured. Given that the K equals 1, the system will try to keep the concentration of both products and reagents the same.

3. A catalyst is a substance that, when added, provides a different and faster mechanism through which a reaction takes place. This only means that the speed at which the equilibrium is attained is reduced, but the enzyme does nothing to alter the difference in energy (ΔG°) of the start and end points of the reaction, which ultimately gives us the value of Keq.

4. The addition of a side reaction does not change the value of Keq for the main reaction. They are both separate ways of making oxaloacetate disappear. While the Keq does not change, keep in mind that the end concentrations will not be the same, for any set of starting concentrations of your substances.

4 0
2 years ago
Analysis of a gaseous chlorofluorocarbon, CClxFy, shows that it contains 11.79% C and 69.57% Cl. In another experiment, you find
uranmaximum [27]

Answer:

The molecular formula = C_2Cl_{4}F_2

Explanation:

Moles =\frac {Given\ mass}{Molar\ mass}

% of C = 11.79

Molar mass of C = 12.0107 g/mol

<u>% moles of C = 11.79 / 12.0107 = 0.9816</u>

% of Cl = 69.57

Molar mass of Cl = 35.453 g/mol

<u>% moles of Cl = 69.57 / 35.453 = 1.9623</u>

Given that the gaseous chlorofluorocarbon only contains chlorine, flourine and carbon. So,

% of F = 100% - % of C - % of C = 100 - 11.79 - 69.57 = 18.64

Molar mass of F = 18.998 g/mol

<u>% moles of F = 18.64 / 18.998 = 0.9812</u>

Taking the simplest ratio for C, Cl and F as:

0.9816 : 1.9623 : 0.9812

= 1 : 2 : 1

The empirical formula is = CCl_2F

Also, Given that:

Pressure = 21.3 mm Hg

Also, P (mm Hg) = P (atm) / 760

Pressure = 21.3 / 760 = 0.02803 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K  

Volume = 458 mL  = 0.458 L (1 mL = 0.001 L)

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.02803 atm × 0.458 L = n × 0.0821 L.atm/K.mol × 298.15 K  

⇒n = 0.00052445 moles

Given that :  

Amount  = 0.107 g  

Molar mass = ?

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.00052445= \frac{0.107\ g}{Molar\ mass}

Molar\ mass= 204.0233\ g/mol

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 1×12.0107 + 2×35.453 + 1×18.998 = 101.9147 g/mol

Molar mass = 204.0233 g/mol

So,  

Molecular mass = n × Empirical mass

204.0233 = n × 101.9147

⇒ n = 2

<u>The molecular formula = C_2Cl_{4}F_2</u>

6 0
3 years ago
8 points
butalik [34]

Answer:

1.3 meters

Explanation: use newton third law equation.

5 0
2 years ago
Is carbon dioxide a iconic or covalent compound
igomit [66]
It is a covalant bond. Because any compound made up of nonmetals will be covalant. Compounds made up of a non metal and metal willl form a ionic compund. :)
3 0
2 years ago
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