Answer:
Both answer choice 2 and answer choice 3 are correct choices.
Explanation:
Biotic factors are living components in an ecosystem. They are living organisms which affect another living component in an ecosystem.
Antibiotic factors are non living components of an ecosystem. They are chemicals which affect living organisms.
The choice 2 is correct. The seeds spread out by mice is a biotic factor interacting with an antibiotic factor
Choice 3 is correct. The seeds in the soil grow into new trees are biotic factors interacting with an antibiotic factor
The answer is D. Heterogeneous mixture
Explanation:
Since 10mm is 1cm
<em>There</em><em>fore</em><em> </em><em>7</em><em>.</em><em>6</em><em>0</em><em>c</em><em>m</em><em> </em><em>is</em><em> </em><em>7</em><em>6</em><em>0</em><em>m</em><em>m</em>
<em>Since</em><em> </em><em>1</em><em> </em><em>inch</em><em> </em><em>is</em><em> </em><em>2</em><em>.</em><em>5</em><em>4</em><em>c</em><em>m</em>
<em>There</em><em>fore</em><em> </em><em>7</em><em>.</em><em>6</em><em>0</em><em>c</em><em>m</em><em> </em><em>is</em><em> </em><em>3</em><em>.</em><em>1</em><em>0</em><em> </em><em>inches</em>
<span>the molar mass of a compound is the sum of the products of the atomic masses by the number of atoms of the element.
molar mass of Na</span>₂SO₄<span> is - 142 g/mol.
1 mol of </span>Na₂SO₄<span> has a mass of 142 g.
In 1 mol of </span>Na₂SO₄<span> the mass of Na is 23 g/mol x 2 = 46 g.
Mass of Na in 1 mol of </span>Na₂SO₄ is - 46 g
mass of Na in 0.820 mol of Na₂SO₄ - 46 g /1 mol x 0.820 mol = 37.72 g.
mass of Na is 37.72 g
<h3>
Answer:</h3>
5.6 L
<h3>
Explanation:</h3>
We are given;
- Initial volume, V1 = 3.5 L
- Initial pressure, P1 = 0.8 atm
- Final pressure, P2 = 0.5 atm
We are required to calculate the final volume;
- According to Boyle's law, the volume of a fixed mass of a gas and the pressure are inversely proportional at a constant temperature.
- That is; P α 1/V
- Mathematically, P=k/V
- At two different pressure and volume;
P1V1 = P2V2
In this case;
Rearranging the formula;
V2 = P1V1 ÷ P2
= (0.8 atm × 3.5 L) ÷ 0.5 atm
= 5.6 L
Therefore, the resulting volume is 5.6 L