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3 years ago

Chemical reaction when chromium metal is immersed in an aqueous solution of cobalt(II) chloride.

Chemistry

1 answer:

Marrrta [24]3 years ago

3 0

$2Cr + 3CoCl_2$ → $2CrCl_3 + 3Co$

Explanation:

The products formed are chromic chloride and cobalt.

Chromium + Cobaltous Chloride = Chromic Chloride + Cobalt

Type of reaction is Single Displacement (Substitution) which is there is a displacement of one atom.

Reactants used in the reaction are -

Chromium $(Cr)$

Cobaltous Chloride $(CoCl_2)$

Products formed in the reaction are -

Chromic Chloride $(CrCl_3)$

Cobalt $(Co)$

Hence, the chemical reaction is as follows -

$Cr + CoCl_2$ → $CrCl_3 + Co$

For balancing the above chemical equation we need to add a coefficient of 2 in front of chromium and of 3 in front of cobalt(II)chloride on right-hand-side while of 2 in front of chromium chloride and of 3 in front of carbon monoxide on left-hand-side of the equation.

Hence, the balanced equation is -

$2Cr + 3CoCl_2$ → $2CrCl_3 + 3Co$

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The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH. It requires 18.

olchik [2.2K]

Answer:

Approximately $0.291\; \rm M$ (rounded to two significant figures.)

Explanation:

The unit of concentration $\rm M$ is the same as $\rm mol \cdot L^{-1}$ (moles per liter.) On the other hand, the volume of both the $\rm NaOH$ solution and the original $\rm HCl$ solution here are in milliliters. Convert these two volumes to liters:

$V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L$ .
$V(\text{$\mathrm{HCl}$, original}) = 10.0\; \rm mL = 10.0\times 10^{-3}\; \rm L = 0.0100\; \rm L$ .

Calculate the number of moles of $\rm NaOH$ in that $0.0182\; \rm L$ of $0.160\; \rm M$ solution:

$\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}$ .

$\rm HCl$ reacts with $\rm NaOH$ at a one-to-one ratio:

$\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l)$ .

Coefficient ratio:

$\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1$ .

In other words, one mole of $\rm NaOH$ would neutralize exactly one mole of $\rm HCl$ . In this titration, $0.291\; \rm mol$ of $\rm NaOH\!$ was required. Therefore, the same amount of $\rm HC$ should be present in the original solution:

$\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}$ .

Calculate the concentration of the original $\rm HCl$ solution:

$\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M$ .

5 0

3 years ago

A bacteria culture begins with 15 bacteria which double in amount at the end of every hour. Solve for the number of bacteria tha

pshichka [43]

The number of bacteria is given by:
N(t) = N(o) x 2ⁿ
Where N(t) is the number after n hours have passed and N(o) is the original number which is 15.
The number grown in the 12th hour is the difference in the number after the 11th and the 12th hour. Thus:
15 x 2¹² - 15 x 2¹¹
= 30,720 bacteria

5 0

3 years ago

What is the final volume?

zaharov [31]

Answer:

Option A. 9.4 L

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 8 L

Initial temperature (T₁) = 293 K

Final temperature (T₂) = 343 K

Final volume (V₂) =?

V₁ / T₁ = V₂ / T₂

8 / 293 = V₂ / 343

Cross multiply

293 × V₂ = 8 × 343

293 × V₂ = 2744

Divide both side by 293

V₂ = 2744 / 293

V₂ = 9.4 L

Therefore, the final volume of the gas is 9.4 L

6 0

2 years ago

How many liters of water can be made from 55 grams of oxygen gas and an excess of hydrogen at a pressure of 12.4 atm and a tempe

polet [3.4K]

First, we need the no.of moles of O2 = mass/molar mass of O2
= 55 g / 32 g/mol
= 1.72 mol
from the balanced equation of the reaction:
2H2 (g) + O2(g) → 2H2O(g)
we can see that the molar ratio between O2: H2O = 1: 2
So we can get the no.of moles of H2O = 2 * moles of O2
= 2 * 1.72 mol
= 3.44 mol
So by substitution by this value in ideal gas formula:
PV = nRT

when P = 12.4 atm & n H2O = 3.44 mol & R= 0.0821 & T = 85 + 273=358K

12.4 atm *V = 3.44 * 0.0821 * 358 = 8.15 L
∴ V ≈ 8.2 L

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3 years ago

State two types of plastics and their uses.

Svetllana [295]

Thermoplastic and thermosetting

thermoplastic:- they are easily molded and extruded into films, fibres and packaging.For eg. PVC

thermosetting:-they are hard and durable and can be used for aircraft parts,tires and auto parts .For eg. phenolic resins.

HOPE THIS HELPS YOU ✌️

7 0

2 years ago

Read 2 more answers