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Bezzdna [24]
2 years ago
5

The the balanced equations to: solutions of lithium sulfate and potassium phosphate are mixed together to make lithium phosphate

and potassium sulfate
Chemistry
1 answer:
nevsk [136]2 years ago
8 0

Answer:

3Li_2SO_4+2K_3PO_4\rightarrow 2Li_3PO_4+3K_2SO_4

Explanation:

Hello!

In this case, when lithium sulfate and potassium phosphate solutions are mixed to yield lithium phosphate and potassium sulfate, the following chemical equation comes up:

Li_2SO_4+K_3PO_4\rightarrow Li_3PO_4+K_2SO_4

However, as atoms are not balanced we apply the inspection method to obtain:

3Li_2SO_4+2K_3PO_4\rightarrow 2Li_3PO_4+3K_2SO_4

Best regards!

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The pH of a solution in which the concentration of H+ is 0.010M will be:
shusha [124]

B: 2

If it's right plz give brainliest lol <3

7 0
2 years ago
A stainless steel knife has a volume of 1.2 cubic centimeters and a mass of 9.0 grams.
bija089 [108]

Answer:

7.5

Explanation:

density = mass / volume

9÷1.2 = 7.5

5 0
2 years ago
Diatomic iodine [I2] decomposes at high temperature to form I atoms according to the reaction I2(g)⇌2I(g), Kc = 0.011 at 1200∘C
umka2103 [35]

Answer:

The concentration of I at equilibrium = 3.3166×10⁻² M

Explanation:

For the equilibrium reaction,

I₂ (g) ⇄ 2I (g)

The expression for Kc for the reaction is:

K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}

Given:

\left[I_2_{Equilibrium} \right] = 0.10 M

Kc = 0.011

Applying in the above formula to find the equilibrium concentration of I as:

0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}

So,

\left[I_{Equilibrium} \right]^2=0.011\times 0.10

\left[I_{Equilibrium} \right]^2=0.0011

\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M

<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>

3 0
3 years ago
the gas left in an used aerosol can is at a pressure of 103 kPa at 25 degrees celsius if this can be thrown into fire what is th
Rainbow [258]
Hello!

The pressure of the gas when it's temperature reaches 928 °C is 3823,36 kPa

To solve that we need to apply Gay-Lussac's Law. It states that the pressure of a gas when the volume is left constant (like in the case of a sealed container like an aerosol can) is proportional to temperature. This is the relationship derived from this law that we use to solve this problem:

P2= \frac{P1}{T1}*T2= \frac{103 kPa}{25}*928=3823,36 kPa

Have a nice day!
5 0
3 years ago
Read 2 more answers
Calculate the hydroxide ion concentration [OH-] for a solution with a pH of 6.10
nalin [4]

There are different formula you need to keep in mind when solving for [OH-]

Given that pH = 6.10

pH + pOH = 14

6.10 + pOH = 14

pOH = 7.9

[OH-] = 10^(-pOH)

[OH-] = 10^(-7.9)

[OH-] = 0.000000013

[OH-] = 1.3 x 10^-8


<h2><u>Answer: [OH-] = 1.3 x 10^-8</u></h2>
4 0
2 years ago
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