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algol [13]
3 years ago
7

9

Chemistry
1 answer:
Svetllana [295]3 years ago
5 0

Answer:

0.0025moles

Explanation:

Molarity of a solution (M) = number of moles (n) ÷ volume (V)

According to this question, to make 250 mL of a 0.01 M solution of CaCl, the following number of moles is needed:

Volume = 250mL = 250/1000 = 0.250Litres.

Using; molarity = n/V

0.01 = n/0.250

n = 0.0025

n = 2.5 × 10^-3 moles.

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A 4.0 L container holds a sample of hydrogen gas at 306 K and 150 kPa. If the pressure increases to 300 kPa and the volume remai
riadik2000 [5.3K]

Answer:

612 K

Explanation:

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

Initial temperature (T₁) = 306 K

Initial pressure (P₁) = 150 kPa

Final pressure (P₂) = 300 kPa

Volume = 4 L = constant

Final temperature (T₂) =?

Since the volume is constant, the final (i.e the new) temperature of the gas can be obtained as follow:

P₁ / T₁ = P₂ / T₂

150 / 306 = 300 / T₂

Cross multiply

150 × T₂ = 306 × 300

150 × T₂ = 91800

Divide both side by 150

T₂ = 91800 / 150

T₂ = 612 K

Thus, the new temperature of the gas is 612 K

4 0
3 years ago
4 HF(g)+SiO2(s) → SiF4(9)+2 H2O(9) <br> Is the Si oxidized or reduced?
Airida [17]

Answer:

Si is reduced since it loses the oxygen atom

8 0
3 years ago
If an object has a volume of 2 millimeters and a mass of 10 grams, calculate the density of the object. (4 points)
kow [346]
Density = g/mL

10 g / 2 mL = 5 g/mL


3 0
3 years ago
Name the following Type 3 compound
Natasha_Volkova [10]

Answer:

32

Explanation:

5 0
3 years ago
Consider the insoluble compound nickel(II) hydroxide , Ni(OH)2 . The nickel ion also forms a complex with cyanide ions . Write a
natali 33 [55]

Answer: Equilibrium constant for this reaction is 2.8 \times 10^{15}.

Explanation:

Chemical reaction equation for the formation of nickel cyanide complex is as follows.

Ni(OH)_{2}(s) + 4CN^{-}(aq) \rightleftharpoons [Ni(CN)_{4}^{2-}](aq) + 2OH^{-}(aq)

We know that,

      K = K_{f} \times K_{sp}

We are given that, K_{f} = 1.0 \times 10^{31}

and,    K_{sp} = 2.8 \times 10^{-16}

Hence, we will calculate the value of K as follows.

     K = K_{f} \times K_{sp}

     K = (1.0 \times 10^{31}) \times (2.8 \times 10^{-16})

        = 2.8 \times 10^{15}

Thus, we can conclude that equilibrium constant for this reaction is 2.8 \times 10^{15}.

4 0
3 years ago
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