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Softa [21]
4 years ago
7

1. The answer to the research question is the...

Chemistry
1 answer:
katovenus [111]4 years ago
8 0

Answer:

1. A claim

2. B evidence

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From goal line to goal line, a football field is 300 ft long. If a player catches the ball while standing on one goal line, runs
KATRIN_1 [288]

Answer:

The player ran 91.44m.

Explanation:

The problem gives you the total distance between goal line to goal line in feet, and the answer must be given in meters, so you should convert the distance the player run from ft to m, because the player run the same distance from goal line to goal line to scores the touchdown.

So, you should apply the following conversion factor:

300ft*\frac{0.3048m}{1ft}= 91.44m

The player ran 91.44m.

3 0
3 years ago
What is the normality of a solution containing 14.8 g of Ca(OH)₂ in 250.0 mL?
aliina [53]

Answer:

Explanation:Are You From Milo?

6 0
3 years ago
What is the mass of 18.0 mL of honey if its' density is 1.42 g/mL?
BigorU [14]

mass (m) = ? , volume = 18.0 ml , density = 1.42 g/ml .

density =  \frac{mass}{volume}  \\  \\ d =  \frac{m}{v}  \\  \\ mass = density \times volume \\  \\ m = d \times v \\  \\ m = 18 \: ml \times  \: 1.42 \:  \frac{g}{ml}  \\  \\ m = 25.56 \: g

I hope I helped you^_^

8 0
3 years ago
Which atom is smaller? (1 point) A. Hydrogen B. Oxygen
frutty [35]
Hydrogen because it only has one electron
6 0
3 years ago
Suppose you want to create a 6 ng/μL solution in a 25 mL volumetric flask. However, this concentration cannot really be accurate
Over [174]

Answer:

Mass of chemical = 1.5 mg

Explanation:

Step 1: First calculate the concentration of the stock solution required to make the final solution.

Using C1V1 = C2V2

C1 = concentration of the stock solution; V1 = volume of stock solution; C2 = concentration of final solution; V2 = volume of final solution

C1 = C2V2/V1

C1 = (6 * 25)/ 0.1

C1 = 1500 ng/μL = 1.5 μg/μL

Step 2: Mass of chemical added:

Mass of sample = concentration * volume

Concentration of stock = 1.5 μg/μL; volume of stock = 10 mL = 10^6 μL

Mass of stock = 1.5 μg/μL * 10^6 μL = 1.5 * 10^6 μg = 1.5 mg

Therefore, mass of sample = 1.5 mg

4 0
3 years ago
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