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Keith_Richards [23]

2 years ago

12

How much heat is absorbed by a 56g iron skillet when its temperature rises from 15oC to 27oC?

1 answer:

wel2 years ago

4 0

we know that $Q=mc_{p}$ Δ $T$

$m=56g$

° $T_{1} =15$ ° $C$

$T_{2} =27$ ° $C$

$Q=56*0.450*(27-15)=302.4J$

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How many formula units are there in 450 g of Na2S04?

Mashutka [201]

Formula units in 450 g of $Na_{2} So_{4}$ is 1.93 × 10²⁴ formula units.

<u>Explanation:</u>

First we have to find the number of moles in the given mass by dividing the mass by its molar mass as,

$$\frac{450 g}{142.04 g/mol} = 3.2 moles$

Now, we have to multiply the number of moles of Na₂SO₄ by the Avogadro's number, 6.022 × 10²³ formula units/mol, so we will get the number of formula units present in the given mass of the compound.

3.2 mol × 6.022 × 10²³ = 1.93 × 10²⁴ formula units.

So, 1.93 × 10²⁴ formula units is present in 450g of Na₂SO₄.

4 0

2 years ago

If you assume this reaction is driven to completion because of the large excess of one ion, what is the concentration of [Fe(SCN

viktelen [127]

Answer : The concentration of $[Fe(SCN)]^{2+}$ is, $4.32\times 10^{-4}M$

Explanation :

When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is $SCN^-$ and $Fe^{3+}$ is excess reagent.

First we have to calculate the moles of KSCN.

$\text{Moles of }KSCN=\text{Concentration of }KSCN\times \text{Volume of solution}$

$\text{Moles of }KSCN=0.00180M\times 0.006L=1.08\times 10^{-5}mol$

Moles of KSCN = Moles of $K^+$ = Moles of $SCN^-$ = $1.08\times 10^{-5}mol$

Now we have to calculate the concentration of $[Fe(SCN)]^{2+}$

$\text{Concentration of }[Fe(SCN)]^{2+}=\frac{\text{Moles of }[Fe(SCN)]^{2+}}{\text{Volume of solution}}$

Total volume of solution = (6.00 + 5.00 + 14.00) = 25.00 mL = 0.025 L

$\text{Concentration of }[Fe(SCN)]^{2+}=\frac{1.08\times 10^{-5}mol}{0.025L}=4.32\times 10^{-4}M$

Thus, the concentration of $[Fe(SCN)]^{2+}$ is, $4.32\times 10^{-4}M$

7 0

2 years ago

Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at 100 °C is ________ atm. (a

svetoff [14.1K]

Answer:

The answer to your question is P = 1.357 atm

Explanation:

Data

Volume = 22.4 L

1 mol

temperature = 100°C

a = 0.211 L² atm

b = 0.0171 L/mol

R = 0.082 atmL/mol°K

Convert temperature to °K

Temperature = 100 + 273

= 373°K

Formula

$(P + \frac{a}{v^{2}} )(v - b) = RT$

Substitution

$(P + \frac{0.211}{22.4})(22.4 - 0.0171) = (0.082)(373)$

Simplify

(P + 0.0094)(22.3829) = 30.586

Solve for P

P + 0.0094 = $\frac{30.586}{22.3829}$

P + 0.0094 = 1.366

P = 1.336 - 0.0094

P = 1.357 atm

7 0

2 years ago

How to seperate sand and water

strojnjashka [21]

Answer: Some kind of strainer system

Explanation:

8 0

2 years ago

Read 2 more answers

Fill in the blanks below.

Elenna [48]

Answer:

Explanation:

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This is the chemical equation given .

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2. In the equation, CO₂ is a___reactant_____.

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6. The number of oxygen atoms on the left side of the equation is__equal to_________ the number of oxygen atoms on the right side.

5 0

2 years ago