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3 years ago

What is the value of c? 56.70

Chemistry

1 answer:

Gennadij [26K]3 years ago

6 0

Due to the lack of information provided, the best I can give you is this:

In Chemistry, the letter c most commonly refers to specific heat, which is present in the ratio J/g°C.

Hope it helps :)

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When 18.0 mL of a 8.43×10-4 M cobalt(II) fluoride solution is combined with 22.0 mL of a 9.72×10-4 M sodium hydroxide solution d

Eva8 [605]

Answer:

Q = 1.08x10⁻¹⁰

Yes, precipitate is formed.

Explanation:

The reaction of CoF₂ with NaOH is:

CoF₂(aq) + 2 NaOH(aq) ⇄ Co(OH)₂(s) + 2 NaF(aq).

The solubility product of the precipitate produced, Co(OH)₂, is:

Co(OH)₂(s) ⇄ Co²⁺(aq) + 2OH⁻(aq)

And Ksp is:

Ksp = 3x10⁻¹⁶= [Co²⁺][OH⁻]²

Molar concentration of both ions is:

[Co²⁺] = 0.018Lₓ (8.43x10⁻⁴mol / L) / (0.018 + 0.022)L = 3.79x10⁻⁴M

[OH⁻] = 0.022Lₓ (9.72x10⁻⁴mol / L) / (0.018 + 0.022)L = 5.35x10⁻⁴M

Reaction quotient under these concentrations is:

Q = [3.79x10⁻⁴M] [5.35x10⁻⁴M]²

Q = 1.08x10⁻¹⁰

As Q > Ksp, the equilibrium will shift to the left producing Co(OH)₂(s) the precipitate

8 0

4 years ago

Pls help 25 points need answer soon thanks

umka21 [38]

4.167 m/per second 250 m/per minute

4 0

3 years ago

What is the partial pressure (in atm) of CO₂ at 468.2 K in a 25.0 L fuel combustion vessel if it contains 60.0 grams CO₂, 82.1 g

NeTakaya

Answer:

2.09 atm

Explanation:

Step 1: Given and required data

Mass of CO₂ (m): 60.0 g

Volume of the vessel (V): 25.0 L

Temperature (T): 468.2 K

We won't need the data of water and uncombusted fuel, since the partial pressures are independent of each other.

Step 2: Calculate the number of moles (n) corresponding to 60.0 g of CO₂

The molar mass of CO₂ is 44.01 g/mol.

60.0 g × 1 mol/44.01 g = 1.36 mol

Step 3: Calculate the partial pressure of CO₂

We will use the ideal gas equation.

P × V = n × R × T

P = n × R × T/ V

P = 1.36 mol × (0.0821 atm.L/mol.K) × 468.2 K/ 25.0 L = 2.09 atm

5 0

3 years ago

2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 2.50 M in concentration.How many mL are required? Plus don'

Morgarella [4.7K]

Answer:

1.36 × 10³ mL of water.

Explanation:

We can utilize the dilution equation. Recall that:

$\displaystyle M_1V_1= M_2V_2$

Where M represents molarity and V represents volume.

Let the initial concentration and unknown volume be M₁ and V₁, respectively. Let the final concentration and required volume be M₂ and V₂, respectively. Solve for V₁:

$\displaystyle \begin{aligned} (2.50\text{ M})V_1 &= (0.800\text{ M})(2.00\text{ L}) \\ \\ V_1 & = 0.640\text{ L} \end{aligned}$

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
$\displaystyle 2.00\text{ L} - 0.640\text{ L} = 1.36\text{ L}$

Convert this value to mL:
$\displaystyle 1.36\text{ L} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 1.36\times 10^3\text{ mL}$

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.

8 0

2 years ago

What is the molecular weight of one mole of H2O?

KatRina [158]

The molecular weight of one mole of H2O is 18.01528(33) g/mol.

8 0

4 years ago