What is not a factor in soil type?

Metallic bonding occurs between atoms of

Pavel [41]

Answer:

Metallic bonding occurs when you have a metal in the solid or liquid state. The s and p valence electrons of metals are loosely held. They leave their “own” metal atoms. This forms a "sea" of electrons that surrounds the metal cations.

Explanation:

4 0

3 years ago

How are h2o and NaCL different

MAVERICK [17]

Answer:

1. H2O is a covalent bond, NaCl is an ionic bond

2. H2O uses hydrogen bonding (H-bonds), NaCl does not

3. H2O is liquid at room temperature. NaCl is solid.

Explanation:

3 0

3 years ago

How many moles of methane (CH4) is<br> 6.70 x 1022 atoms of methane?

suter [353]

Answer:

0.1113 mol

Explanation:

Data Given:

no. of atoms of CH₄= 6.70 x 10²² atoms

no. of moles of methane (CH₄) = ?

Solution:

we will find no. of moles of methane (CH₄)

Formula used

no. of moles = no. of atoms / Avogadro's number

Where

Avogadro's number = 6.022 x 10²³

Put values in above equation

no. of moles = 6.70 x 10²² atoms / 6.022 x 10²³ (atoms/mol)

no. of moles = 0.1113 mol

So,

There are 0.1113 moles of methane.

8 0

3 years ago

When a 3.00 grams sample of a compound containing only c, h, and o was completely burned, 1.17 grams of h2o and 2.87 grams of co

SVEN [57.7K]

Answer:- $CH_2O_2$

Solution:- From given masses of carbon dioxide and water we could calculate the moles that helps to calculate the moles of C and H.

Molar mass of carbon dioxide = 44 gram per mol

molar mass of water = 18.02 gram per mol

From given info, combustion of compound gives 1.17 grams of water and 2.87 grams of carbon dioxide. Let's calculate the moles of these:

$1.17gH_2O(\frac{1mol}{18.02g})$

= $0.0649molH_2O$

Similarly, $2.87gCO_2(\frac{1mol}{44g})$

= $0.0652molCO_2$

One mol of water has two moles of H. So, the moles of H would be two times the moles of water as calculated above.

So, moles of H = 2* 0.0649 = 0.1298 mol

One mol of carbon dioxide contains one mol of C. So, the moles of C would be equal to the moles of carbon dioxide calculated above.

moles of C = 0.0652 mol

Let's convert the moles of H and C to grams so that we could calculate the amount of oxygen present in the sample as:

grams of H in sample = 1.008 x 0.1298 = 0.1308 g

grams of C in sample = 12*0.0652 = 0.7824 g

If we subtract the sum of the masses of C and H from sample mass then it would give as the mass of oxygen since the sample has only C, H and O.

mass of O in sample = 3.00g - (0.1308 g + 0.7824 g)

= 3.00 g - 0.9132 g

= 2.0868 g

Let's convert these grams of oxygen to moles on dividing by it's atomic mass as:

$2.0868gO(\frac{1mol}{15.999g})$

= 0.130 mol O

Now, we have the moles of all the three atoms and we know that an empirical formula is the simplest whole number ratio of the moles of atoms. So, let's calculate the ratio. For this, we divide the moles of each by the least one of them.Looking at the moles, the least value is for carbon. So, let's divide the moles of each by the moles of C as:

C = $\frac{0.0652}{0.0652}$ = 1