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Viefleur [7K]
2 years ago
5

Which image best represents a solution of distilled water at pH = 7?

Chemistry
1 answer:
lara31 [8.8K]2 years ago
5 0

Answer:

nuetral

Explanation:

water has neutral

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KengaRu [80]
Both are oxidation reactions.  Burning is just a lot faster than rusting.
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What determines a molecules VSEPR shape?
frosja888 [35]
Using the VSEPR theory, the electron bond pairs and lone pairs on the center atom will help us predict the shape of a molecule. The shape of a molecule is determined by the location of the nuclei and its electrons. The electrons and the nuclei settle into positions that minimize repulsion and maximize attraction.
3 0
3 years ago
Sample A: 300 mL of 1M sodium chloride
yarga [219]

Answer:

sample B contains the larger density

Explanation:

Given;

volume of sample A, V = 300 mL = 0.3 L

Molarity of sample A, C = 1 M

volume of sample B, V = 145 mL = 0.145 L

Molarity of sample B, C = 1.5 M

molecular mass of sodium chloride, Nacl = 23 + 35.5 = 58.5 g/mol

Molarity is given as;

C = \frac{moles \ of \ solute, \ mol}{liters \ of \ solvent} \\\\Moles \ of \ solute \ for \ sample \ A = 1 \times 0.3 = 0.3 \ mol\\\\Moles \ of \ solute \ for \ sample \ B = 1.5 \times 0.145 = 0.2175 \ mol

The reacting mass for sample A = 0.3mol x  58.5 g/mol = 17.55 g

The reacting mass for sample B = 0.2175 mol x 58.5 g/mol = 12.72 g

The density of sample A  = \frac{mass}{volume} = \frac{17.55}{0.3} = 58.5 \ g/L

The density of sample B = \frac{mass}{volume} = \frac{12.72}{0.145} = 87.72 \ g/L

Therefore, sample B contains the larger density

5 0
3 years ago
What volume of Co2 (carbon (iv) oxide)
hram777 [196]

Answer:

2.1056L or 2105.6mL

Explanation:

We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:

Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol

Mass of Na2CO3 = 10g

Mole of Na2CO3 =.?

Mole = mass /molar mass

Mole of Na2CO3 = 10/106

Mole of Na2CO3 = 0.094 mole

Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:

Na2CO3 + 2HCl —> 2NaCl + H2O + CO2

From the balanced equation above,

1 mole of Na2CO3 reacted to produce 1 mole of CO2.

Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.

Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:

1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.

Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L

Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL

7 0
4 years ago
1. Explain the two processes that keep atmospheric oxygen and carbon dioxide at stable levels?
lesya [120]

Cellular Respiration and Photosynthesis. Photosynthesis is the process of when plants use sunlight to make foods from carbon dioxide and water, to later on make oxygen. Cellular respiration is the process through which cells convert sugars into energy.

4 0
3 years ago
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