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olasank [31]
3 years ago
7

Question: Soil formation begins with the weathering of _________________?

Chemistry
1 answer:
Igoryamba3 years ago
5 0

Answer:

the earth's rocks

Explanation:

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Pollen is produced in the anthers, which are the _______ organs of flowers.
scoray [572]
D, since the anther is a male organ
8 0
3 years ago
A 1.0 mol sample of which of the following compounds has the greatest mass?
Sholpan [36]

Answer:

D) N2O5

Explanation:

The molar mass of a substance is defined as the mass of this substance in 1 mol. To solve this question we must find the molar mass of each option:

<em>Molar mass NO:</em>

1N = 14g/mol*1

1O = 16g/mol*1

14+16 = 30g/mol

<em>Molar mass NO2:</em>

1N = 14g/mol*1

2O = 16g/mol*2

14+32 = 46g/mol

<em>Molar mass N2O:</em>

2N = 14g/mol*2

1O = 16g/mol*1

28+16 = 44g/mol

<em>Molar mass N2O5:</em>

2N = 14g/mol*2

5O = 16g/mol*5

28+80 = 108g/mol

That means the compound with the greatest mass is:

<h3>D) N2O5</h3>
5 0
3 years ago
True or false? Covalent molecules are formed when non-metal atoms share electrons with other non-metal atoms.
Tomtit [17]

Answer:

true

covalent bonds are between non metals and nonmetals. and they are sharing electrons.

7 0
3 years ago
2.0 mol of gas occupy 44.8 l at a temperature of 0°c. what is the pressure of the gas
MArishka [77]

The pressure of the gas is 1.0 bar.

<em>pV</em> = <em>nRT</em>

<em>T</em> = (0 + 273.15) K = 273.15 K

<em>p</em> = (<em>nRT</em>)/<em>V</em> = (2.0 mol × 0.083 14 bar·L·K⁻¹mol⁻¹ × 273.15 K)/44.8 L = 1.0 bar

5 0
3 years ago
A student took a calibrated 250.0 gram mass, weighed it on a laboratory balance, and found it read 266.5 g. What was the student
marta [7]

Answer:

\boxed{\text{6.60 \%}}

Explanation:

\text{Percent error} = \dfrac{\lvert \text{Measured - Actual}\lvert}{ \text{Actual}} \times100 \,\%

\text{\% error} = \dfrac{\lvert 266.5 - 250.0\lvert}{250.0} \times 100 \, \% = \dfrac{\lvert 16.5\lvert}{250.0} \times 100 \, \% = \dfrac{16.5}{250.0} \times 100 \,\%\\\\ = 0.0660 \times 100 \, \% = \textbf{6.60 \%}\\\text{The student's percent error is $\boxed{\textbf{6.60 \%}}$}

4 0
3 years ago
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