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2 years ago

Please Help Me Balance The equation!! will mark brainliest

Chemistry

2 answers:

Delvig [45]2 years ago

8 0

Answer:

XCl2 + 2 AgNO3 = X(NO3)2 + 2 AgCl

Explanation:

i ran this through a calculator

Lina20 [59]2 years ago

5 0

Answer: XCl2(aq) + 2 AgNO3(aq) ---> X(NO3)2(aq) + 2 AgCl(s)

Explanation:

Hehe I remembered how...

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Many homeowners treat their lawns with CaCO3(s) to reduce the acidity of the soil. Write a net ionic equation for the reaction o

Aliun [14]

Many homeowners treat their lawns with CaCO3(s) to reduce the acidity of the soil. The net ionic equation for the reaction of CaCO3(s) with a strong acid, HCl (I chose HCl because it is a strong acid) is CaCO3(s) +2 HCl(aq) → CaCl2(s) + H2O(aq) + CO2(g).

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3 years ago

To which group of the periodic table do lithium and potassium belong?

s344n2d4d5 [400]

Answer:

1)alkali metals

2)1g/ml

3) mercury

Explanation:

HEY!!^^

have a good day

thank me later

carryonlearing

6 0

2 years ago

Read 2 more answers

How many significant figures are in 120 miles?

balu736 [363]

2 significant zeros.

1 and 2 are the significant zeros.

pls mark brainliest

4 0

3 years ago

The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42

denis23 [38]

Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

$k = Ae^{\frac{-Ea}{RT}}$

Where:

k: is the rate constant

A: is the frequency factor

Ea: is the activation energy

R: is the gas constant = 8.314 J/(K*mol)

T: is the temperature

We have for the uncatalyzed reaction:

Ea₁ = 70 kJ/mol

And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

T₂ = 20 °C = 293 K

The frequency factor A is constant and the initial concentrations are the same.

Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

$k_{1} = k_{2}$

$Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}$ (1)

By solving equation (1) for T₁ we have:

$T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C$

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!

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3 years ago

True or False? SLOW moving water deposits sediment.

wolverine [178]

no probs at all :)

5 pts

7 0

3 years ago