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mrs_skeptik [129]
3 years ago
8

What is the physical state of each of these materials at room temperature?

Chemistry
2 answers:
zloy xaker [14]3 years ago
5 0
At room temperature:
a. Gold is a solid, as in the gold jewelry that most people have.
b. Gasoline is a liquid, as it is pumped in cars to use as fuel.
c. Oxygen is a gas, as we breathe it in from the atmosphere.
d. Neon is a gas, it is one of the noble gases in the periodic table.
e. Olive oil is a liquid, as it is used in cooking.
f. Sulfur is a solid, and it can be found naturally in mining.
g. Mercury is a liquid which is often seen in thermometers.
Paladinen [302]3 years ago
4 0
<span>a. solid
b. liquid
c. gas
d. solid
e. liquid
f. solid
g. liquid

Matter and substance are definitely related. All substances are matter but all matters are not substance. A matter can consist of numerous substances. Matter is generally a loose term used in respect to a substance. Any physical object can be casually called a matter. Matter and substance are sometimes used for the same context, but it is completely wrong. Numerous examples have already proved that a matter may or may not be a substance depending on its physical nature, but a substance is always a matter.<span>
</span></span>
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Answer:

it moves faster when heated

5 0
3 years ago
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The compound borazine consists of 40.29% boron, 7.51%
ioda

Answer:

B_3H_6N_3

Explanation:

Hello there!

In this case, since the mass percentages in a compound which is wanted to know the molecular formula, can be assumed to be the masses, we first need to compute the moles they have in the formula unit:

n_B=40.29gB*\frac{1molB}{10.811gB} =3.73molB\\\\n_H=7.51gH*\frac{1molH}{1.01gH} =7.44molH\\\\n_N=52.20gN*\frac{1molN}{14.01gN} =3.73molN

Next, we divide each moles by the fewest ones (3.73 mol) in order to find the subscript in the empirical formula first:

B:\frac{3.73}{3.73}=1 \\\\H:\frac{7.44}{3.73}=2\\\\N:\frac{3.73}{3.73}=1

Then, the empirical formula is BH2N whose molar mass is 26.83 g/mol, so the ratio of molecular to empirical is 80.50/26.83=3; therefore, the molecular formula is three times the empirical one:

B_3H_6N_3

Best regards!

7 0
3 years ago
What is the entropy of this collection of training examples with respect to the positive class B. What are the information gains
Alenkinab [10]

The data set is missing in the question. The data set is given in the attachment.

Solution :

a). In the table, there are four positive examples and give number of negative examples.

Therefore,

$P(+) = \frac{4}{9}$   and

$P(-) = \frac{5}{9}$

The entropy of the training examples is given by :

$ -\frac{4}{9}\log_2\left(\frac{4}{9}\right)-\frac{5}{9}\log_2\left(\frac{5}{9}\right)$

= 0.9911

b). For the attribute all the associating increments and the probability are :

  $a_1$   +   -

  T   3    1

  F    1    4

Th entropy for   $a_1$  is given by :

$\frac{4}{9}[ -\frac{3}{4}\log\left(\frac{3}{4}\right)-\frac{1}{4}\log\left(\frac{1}{4}\right)]+\frac{5}{9}[ -\frac{1}{5}\log\left(\frac{1}{5}\right)-\frac{4}{5}\log\left(\frac{4}{5}\right)]$

= 0.7616

Therefore, the information gain for $a_1$  is

  0.9911 - 0.7616 = 0.2294

Similarly for the attribute $a_2$  the associating counts and the probabilities are :

  $a_2$  +   -

  T   2    3

  F   2    2

Th entropy for   $a_2$ is given by :

$\frac{5}{9}[ -\frac{2}{5}\log\left(\frac{2}{5}\right)-\frac{3}{5}\log\left(\frac{3}{5}\right)]+\frac{4}{9}[ -\frac{2}{4}\log\left(\frac{2}{4}\right)-\frac{2}{4}\log\left(\frac{2}{4}\right)]$

= 0.9839

Therefore, the information gain for $a_2$ is

  0.9911 - 0.9839 = 0.0072

$a_3$     Class label      split point       entropy        Info gain

1.0         +                        2.0            0.8484        0.1427

3.0        -                         3.5            0.9885        0.0026

4.0        +                        4.5            0.9183         0.0728

5.0        -

5.0        -                        5.5            0.9839        0.0072

6.0        +                       6.5             0.9728       0.0183

7.0        +

7.0        -                        7.5             0.8889       0.1022

The best split for $a_3$  observed at split point which is equal to 2.

c). From the table mention in part (b) of the information gain, we can say that $a_1$ produces the best split.

4 0
3 years ago
If a 1.45 M solution has 2.43 g HCl dissolved, what is the volume of solution? (Change g HCl into mol using molar mass)
Ulleksa [173]

Answer: The volume of solution is 0.0459 L

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution

Molarity=\frac{n}{V_s}

where,

n = moles of solute

V_s = volume of solution in L

moles of HCl (solute) = \frac{\text {given mass}}{\text {Molar mass}}=\frac{2.43g}{36.5g/mol}=0.0666

Now put all the given values in the formula of molality, we get

1.45M=\frac{0.0666}{V_s}

V_s=\frac{0.0666}{1.45}=0.0459

Therefore, the volume of solution is 0.0459 L

5 0
3 years ago
How many atoms are in a sample of 68.7 g copper (Cu)?
dlinn [17]
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8 0
4 years ago
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