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Studentka2010 [4]
3 years ago
15

Elements can be described by various properties, and identified by their boiling and melting points. For example, gold melts at

1,064ºC and boils at 2,856ºC. Does boiling point depend on the mass present?
A. No; chemical properties stay the same regardless of mass.
B. No; physical properties stay the same regardless of mass.
C. Yes; physical properties can change when mass increases or decreases.
D. No; qualitative properties like boiling point stay the same regardless of mass.
Chemistry
2 answers:
jekas [21]3 years ago
7 0

Answer:

B i supposed.

Explanation:

aleksandr82 [10.1K]3 years ago
5 0

Answer:

Explanation:

Melting and boiling point variations are not clear (do not have uniform pattern) in periodic table. But we can see, some elements have higher melting and boiling points and some have less. Here we study melting and boiling points of s, p, d blocks elements. IVAth group elements (C,Si) show high melting and boiling points because they have covalent gigantic lattice structures.

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Burning 12.00 g of an oxoacid produces 17.95 g of carbon dioxide and 4.87 g of water. Consider that 0.25
Veseljchak [2.6K]

Answer: The molecular formula will be C_6H_6O_6

Explanation:

Mass of CO_2 = 17.95 g

Mass of H_2O= 4.87 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 17.95 g of carbon dioxide, =\frac{12}{44}\times 17.95=4.89g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.87 g of water, =\frac{2}{18}\times 4.87=0.541g of hydrogen will be contained.

Mass of oxygen in the compound = (12.00) - (4.89+0.541) = 6.57 g

Mass of C = 4.89 g

Mass of H =  0.541 g

Mass of O = 6.57 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{4.89g}{12g/mole}=0.407moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.541g}{1g/mole}=0.541moles

Moles of O=\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{6.57g}{16g/mole}=0.410moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.407}{0.407}=1

For H =\frac{0.541}{0.407}=1

For O = \frac{0.410}{0.407}=1

The ratio of C : H : O = 1: 1  : 1

Hence the empirical formula is CHO.

Hence the empirical formula is CHO

The empirical weight of CHO = 1(12)+1(1)+1(16)= 29 g.

If 0.25 moles has mass of 44.0 g

Thus 1 mole has mass of = \frac{44.0}{0.25}\times 1=176g

Thus molecular mass is 176 g

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{176g}{29g}=6

The molecular formula will be=6\times CHO=C_6H_6O_6

5 0
3 years ago
515282 quarts into milliliters. I
Minchanka [31]

Answer:

487638640,7819

dndndnndndbxbdbdbdbdb

6 0
3 years ago
Why does the earth experience unequal heating
evablogger [386]
Global Warming is tthe cause
3 0
3 years ago
The equilibrium constant for the reaction 2NO2(g) N2O4(g) is Keq = . If a sample at equilibrium was found to contain 0.058 M NO2
Sladkaya [172]

Answer:

\boxed{3.6}

Explanation:

                  2NO₂ ⇌ N₂O₄

E/mol·L⁻¹:   0.058     0.012

K_{\text{eq}} = \dfrac{\text{[N$_{2}$O$_{4}$]}}{\text{[NO$_{2}$]$^{2}$}} = \dfrac{0.012}{0.058^{2}} = \mathbf{3.6} \\\\

\text{The $K_{\text{eq}}$ value would be $\boxed{\mathbf{3.6}}$}

7 0
4 years ago
Match each term to its definition. 1. compound 2. atom 3. element 4. molecule a. the basic unit of matter b. a substance contain
Ksivusya [100]

Answer:

1. C

2. A

3. B

4. D

i think

7 0
3 years ago
Read 2 more answers
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