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Veronika [31]
3 years ago
12

Need a lil help here

Chemistry
2 answers:
Step2247 [10]3 years ago
7 0
OKKKKKKKKKKKKKKKKKKKKKK

harina [27]3 years ago
5 0

Answer:

im not the realest igga is the anwer

Explanation:

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What is the ionic equation of Aqueous calcium bromide was mixed with aqueous gold(I) perchlorate, and a crystallized gold(I) bro
Bumek [7]

Answer:- 2AuClO_4(aq) + CaBr_2(aq)\rightarrow 2AuBr(s)+ Ca(ClO_4)_2(aq)

Explanations:- It's a double replacement reaction where a precipitate of silver(I)bromide is formed. A dpuble replacement reaction in general looks as AB + CD \rightarrow AD + CB

To balance the equation we need to multiply gold compounds on both sides by 2 and the balanced equation is..

2AuClO_4(aq) + CaBr_2(aq)\rightarrow 2AuBr(s)+ Ca(ClO_4)_2(aq)

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3 years ago
I need help in science
astraxan [27]
The answer is D . I hope this help you :) .
7 0
2 years ago
3.2 moles of H3PO4 to grams
Nimfa-mama [501]

Answer:

313, 6grams of H3PO4

Explanation:

We calculate the weight of 1 mol of H3PO4:

Weight 1 mol H3PO4= (Weight H)x3+ (Weight P)+(Weight 0)x4  =1gx3+31g+16gx4

Weight 1 mol H3PO4=98 g /mol

1 mol-----98 grams H3PO4

3,2mol----x= (3,2molx 98 grams H3PO4)/ 1mol=313,6 grams H3PO4

4 0
3 years ago
Which situation is the best example of interia
AveGali [126]
One's body movement to the side when a car makes a sharp turn. Tightening of seat belts in a car when it stops quickly. A ball rolling down a hill will continue to roll unless friction or another force stops it.
6 0
3 years ago
The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the r
rodikova [14]

Answer:

E₁ ≅ 28.96 kJ/mol

Explanation:

Given that:

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,

Let the activation energy for a catalyzed biochemical reaction = E₁

E₁ = ??? (unknown)

Let the activation energy for an uncatalyzed biochemical reaction = E₂

E₂ = 50.0 kJ/mol

    = 50,000 J/mol

Temperature (T) = 37°C

= (37+273.15)K

= 310.15K

Rate constant (R) = 8.314 J/mol/k

Also, let the constant rate for the catalyzed biochemical reaction = K₁

let the constant rate for the uncatalyzed biochemical reaction = K₂

If the  rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:

K₁ = 3.50 × 10³

K₂ = 1

Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;

we can use the formula for Arrhenius equation;

K=Ae^{\frac{-E}{RT}}

If K_1=Ae^{\frac{-E_1}{RT}} -------equation 1     &

K_2=Ae^{\frac{-E_2}{RT}} -------equation 2

\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}

E_1= E_2-RT*In(\frac{K_1}{K_2})

E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})

E_1 = 28957.39292  J/mol

E₁ ≅ 28.96 kJ/mol

∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol

8 0
2 years ago
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