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3 years ago

Given the equation: Ca(CO3) + 2NaOH -----> Ca(OH)2 + Na2(CO3)

Chemistry

1 answer:

nexus9112 [7]3 years ago

3 0

Answer:

there needs to be 4 Ca(CO3)

Explanation:

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What traits do all organisms with no internal themoregulation have?

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Ectothermy and Endothermy

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A weight of 22.25g was obtained when 10-mL of water at 27C was pipet to empty Erlenmeyer flask. What is the volume delivered by

Nimfa-mama [501]

Answer:

the volume delivered by the pipette = 22.32 mL

Explanation:

To calculate this, let us first note that the density of water relates it weight and its volume (density = mass ÷ volume), hence we are going to use density to determine the volume.

Density of water = mass/volume = 0.997 g/ mL

mass = 22.25g

Density = 0.997g/mL

volume = ?

$density = \frac{mass}{volume}\\\therefore volume = \frac{mass}{density}\\volume = \frac{22.25}{0.997}\\volume = 22.32\ mL$

∴ the volume delivered by the pipette = 22.32 mL

Please note that this calculation is based on the fact that the weight of the empty flask has been determined and canceled out.

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3 years ago

If the mass of an object increases, the force acting on it, such as gravitational force, also increases.

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Gravitational force will increase with greater mass

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At equilibrium, the concentrations of the products and reactants for the reaction, H2 (g) + I2 (g)  2 HI (g), are [H2] = 0.106

lana [24]

Answer:

The new equilibrium concentration of HI: [HI] = 3.589 M

Explanation:

Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M

Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M

Given chemical reaction: H₂(g) + I₂(g) → 2 HI(g)

The equilibrium constant ( $K_{c}$ ) for the given chemical reaction, is given by the equation:

$K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}$

At the original equilibrium state:

$K_{c} = \frac {(1.29\: M)^{2}}{(0.106\: M) \times (0.022\: M)}$

$K_{c} = \frac {1.6641}{0.002332} = 713.59$

Therefore, at the new equilibrium state:

$K_{c} = \frac {[HI]^{2}}{(0.95\: M) \times (0.019\: M)}$

$\Rightarrow K_{c} = 713.59 = \frac {[HI]^{2}}{0.01805}$

$\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88$

$\Rightarrow [HI] = \sqrt {12.88} = 3.589 M$

Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M

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3 years ago

Write the electronic configuration of chromium (at no. 24) and copper (at. no. 29).

myrzilka [38]

Answer: Chromium 2)8)8)6

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Explanation:

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