All categories

2 years ago

what rule can you state about the relationship between phase changes and temperature?Between phase changes and heat energy

1 answer:

5 0

The phase does not change until the substance is at the same temperature. Hope this helps! By the way, this was the answer for the question "What rule can you state about the relationship between phase changes and temperature"

You might be interested in

Paintball is a popular recreational activity that uses a metal tank of compressed carbon

Dovator [93]

Answer:

Explanation:

Here we have the mass of CO₂ added = 340 g

From

$Number \, of \, moles = \frac{Mass}{Molar \, mass}$

We have, where the molar mass of CO₂ is 44.01 g/mol

Therefore,

$Number \, of \, moles = \frac{340}{44.01} = 7.73 \, \, \, moles$

71. Included drawing attached

72. Here we have the pressure of the gas given by Charles law which can be resented as follows;

$\frac{P_1}{P_2} =\frac{T_1}{T_2}$

Where:

P₁ = Initial pressure = 6.1 atmospheres

P₂ = Final pressure

T₁ = Initial Temperature = 293 K

T₂ = Initial Temperature = 313 K

Therefore,

$P_2= T_2 \times \frac{P_1}{T_1} = 313 \times \frac{6.1}{293} = 3.312 \, \, \, atmospheres$

5 0

2 years ago

In the synthesis of barium carbonate from an alkali metal carbonate (M2CO3 where M is one of the alkali metals) a student genera

Paraphin [41]

Answer:

0.019 moles of M2CO3

Explanation:

M2CO3(aq) + BaCl2 (aq) --> 2MCl (aq) + BaCO3(s)

From the equation above;

1 mol of M2CO3 reacts to produce 1 mol of BaCO3

Mass of BaCO3 formed = 3.7g

Molar mass of BaCO3 = 197.34g/mol

Number of moles = Mass / Molar mass = 3.7 / 197.34 = 0.0187 ≈ 0.019mol

Since 1 mol of M2CO3 reacts with 1 mol of BaCO3,

1 = 1

x = 0.019

x = 0.019 moles of M2CO3

3 0

2 years ago

Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown c

Alexus [3.1K]

Answer:

0.3229 M HBr(aq)

0.08436M H₂SO₄(aq)

Explanation:

Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.

Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).

NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)

When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.

$18.45 \times 10^{-3} L NaOH.\frac{0.3500molNaOH}{1LNaOH} .\frac{1molHBr}{1molNaOH} .\frac{1}{20.00 \times 10^{-3} LHBr} =\frac{0.3229molHBr}{1LHBr} =0.3229M$

Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?

Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).

2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)

When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.

$42.18 \times 10^{-3} LNaOH.\frac{0.1000molNaOH}{1LNaOH} .\frac{1molH_{2}SO_{4}}{2molNaOH} .\frac{1}{25.00\times 10^{-3}LH_{2}SO_{4}} =\frac{0.08436molH_{2}SO_{4}}{1LH_{2}SO_{4}} =0.08436M$

6 0

2 years ago

A solution is prepared by adding 6.24 g of benzene (C 6H 6, 78.11 g/mol) to 80.74 g of cyclohexane (C 6H 12, 84.16 g/mol). Calcu

tekilochka [14]

Answer:

$x_B=0.0769$

$m=0.990m$

Explanation:

Hello,

In this case, we can compute the mole fraction of benzene by using the following formula:

$x_B=\frac{n_B}{n_B+n_C}$

Whereas n accounts for the moles of each substance, thus, we compute them by using molar mass of benzene and cyclohexane:

$n_B=6.24g*\frac{1mol}{78.11g}=0.0799mol\\ \\n_C=80.74g*\frac{1mol}{84.16g} =0.959mol$

Thus, we compute the mole fraction:

$x_B=\frac{0.0799mol}{0.0799mol+0.959mol}\\ \\x_B=0.0769$

Next, for the molality, we define it as:

$m=\frac{n_B}{m_C}$

Whereas we also use the moles of benzene but rather than the moles of cyclohexane, its mass in kilograms (0.08074 kg), thus, we obtain:

$m=\frac{0.0799mol}{0.08074kg}=0.990mol/kg$

Or just 0.990 m in molal units (mol/kg).

Best regards.

7 0

2 years ago

How many molecules are in 79g of fe2o3

horrorfan [7]

Convert grams —> mols and then mols —> atoms

We know that there are 6.02 x 10^23 atoms/mol

And we know that there are about 160 grams of fe2o3 per mol

So (79g fe2o3)/(160 g/mol) = .49 mol fe2o3

Now we use avogadro’s number to do

(.49 mol fe2o3)/(6.02 x 10^23 atoms/mol) = the answer.

I’ll leave the easy math to you.

7 0

2 years ago