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7nadin3 [17]
3 years ago
5

NEED ANASWER ASAP

Chemistry
1 answer:
kenny6666 [7]3 years ago
8 0
The answer is A ) available food
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Which statement is correct about the rate of a chemical reaction? (5 points)
velikii [3]

Answer:

it does not depend on the temperature

6 0
3 years ago
The Ka of a certain indicator is 4 ×10−7. The color of HIn is green and that of In− is red. A few drops of the indicator are add
Monica [59]

Answer:

7.4 - 5.4.

Explanation:

The relationship between HIn that is the non-ionized form and the In^-; the ionized form is an equilibrium Reaction which is given below;

HIn(aq) + H2O(l) <--------> H3O^+(aq). + In^-(aq).

From the question, we are given that the ka is 4 ×10^−7, therefore, the mathematical relationship between HIn and In^- is given below;

Ka = [H3O^+] [In^-] / [HIn] = 4 ×10^−7.

The formula we are going to be using to solve this question is given below;

pH = pKa (+ or -) 1.

Recall that the relationship between the ka and pKa;

pKa = - log (ka).

pKa = - log ( 4 ×10^−7).

pKa = 6.4

Therefore, for it to have a distinct color change, the pH range will be;

pH = pKa (+ or -) 1.

pH= 6.4 + 1 = 7.4.

pH= 6.4 - 1 = 5.4.

Hence, the pH range = 7.4 - 5.4.

4 0
3 years ago
a glass container was initially charged with 1.50 mol of a gas sample at 3.75 atm and 21.7C. some of the gas was release as the
butalik [34]

Answer:

0.39 mol

Explanation:

Considering the 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

At same volume, for two situations, the above equation can be written as:-

\frac {{n_1}\times {T_1}}{P_1}=\frac {{n_2}\times {T_2}}{P_2}

Given ,  

n₁ = 1.50 mol

n₂ = ?

P₁ = 3.75 atm

P₂ = 0.998 atm

T₁ = 21.7  ºC

T₂ = 28.1 ºC

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

T(K) = T( °C) + 273.15  

So,  

T₁ = (21.7 + 273.15) K = 294.85 K  

T₂ = (28.1 + 273.15) K = 301.25 K  

Using above equation as:

\frac{{n_1}\times {T_1}}{P_1}=\frac{{n_2}\times {T_2}}{P_2}

\frac{{1.50\ mol}\times {294.85\ K }}{3.75\ atm}=\frac{{n_2}\times {301.25\ K  }}{0.998\ atm}

n_2=\frac{{1.50}\times {294.85}\times 0.998}{3.75\times 301.25}\ mol

Solving for n₂ , we get:

n₂ = 0.39 mol

7 0
3 years ago
The following ionic compounds are found in common household products. Name each of the compounds:
Natali5045456 [20]
A) Calcium Dihydrogen Phosphate
B) Iron(II) Sulfate
C) Calcium Carbonate
D) Magnesium Oxide
E) Sodium Nitrite
F) Potassium iodide
7 0
3 years ago
Explain why there might be a change in the density of a forged product as compared to that of the cast blank.
kkurt [141]

Answer:

Forged parts are often tougher than cast parts. This can be determined by performing tensile tests on various areas on the parts. Additionally, the microstructures of forged and cast parts can be used to determine if a part was forged or cast. The microstructure of a cast part will have a more uniform grain structure.

Explanation:

4 0
2 years ago
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