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4 years ago

Which dog would likely have the fastest reaction time? A. a large dog B. a medium-size dog C. a small dog D. a sick dog

Chemistry

2 answers:

denpristay [2]4 years ago

7 0

Answer:

a small dog

Explanation:

Less mass to move

alexandr1967 [171]4 years ago

6 0

A medium sized dog. They are going to have more muscle to run and react. I have a pitbull and let me tell you he is very fast reacting!

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Determine % yield if a student obtains 45 g of product in an experiment and the theoretical amount is determined to be 50 g. *

ycow [4]

Answer:

90%

Explanation:

Percentage yield = ?

Theoretical yield = 50g

Actual yield = 45g

To calculate the percentage yield of a compound, we'll have to use the formula of percentage yield which is the ratio between the actual yield to theoretical multiplied by 100

Percentage yield = (actual yield / theoretical yield) × 100

Percentage yield = (45 / 50) × 100

Percentage yield = 0.9 × 100

Percentage yield = 90%

The percentage yield of the substance is 90%

6 0

3 years ago

What is present in the glucose molecule that is missing in the carbon dioxide molecule?

rusak2 [61]

Hydrogen .carbon dioxide is CO2 and glucose is C6H12O6

3 0

4 years ago

What is the barycenter of the moon and the earth

kirill [66]

Answer:

The barycenter is the point in space around which two objects orbit. For the Moon and Earth, that point is about 1000 miles (1700 km) beneath your feet, or about three-quarters of the way from the Earth's center to its surface.

Explanation:

6 0

3 years ago

Here are some data from a similar experiment, to determine the empirical formula of an oxide of tin. Calculate the empirical for

eduard

Answer:

Empirical formula of the Tin oxide sample is SnO₂

Explanation:

Tin reacts with combines with oxygen to form an oxide of tin.

Mass of crucible with cover = 19.66 g

Mass of crucible, cover, and tin sample = 22.29 g

Mass of crucible and cover and sample, after prolonged heating gives constant weight = 21.76 g

Mass of Tin oxide sample = 22.29 - 19.66 = 2.63 g

Mass of ordinary tin, after heating to breakdown the tin and oxygen = 21.76 - 19.66 = 2.1 g

Meaning that, mass of oxygen in the tin oxide sample = 2.63 - 2.1 = 0.53 g

Mass of Tin in the Tin Oxide sample = 2.1 g

Mass of Oxygen in the Tin oxide sample = 0.53 g

Convert these to number of moles

Number of moles of Tin on the Tin oxide sample = 2.1/118.71 = 0.0177

Number of moles of Oxygen in the Tin oxide sample = 0.53/16 = 0.0335

divide the number of moles by the lowest number

0.0177:0.0335

It becomes,

1:2

SnO₂

Hence, the empirical formula for the Tin oxide sample = SnO₂

7 0

4 years ago

A syringe contains 0.65 moles of he gas that occupy 900.0 ml. what volume (in l) of gas will the syringe hold if 0.35 moles of n

Grace [21]

PV=nRT= a universal constant For any condition P1V1/n1T1=R and P2V2/n2T2=R i.e P1V1/n1T1=P2V2/n2T2 Becomes V1/n1=V2/n2 rearranging and solving V2=V1X(n2/n1)= 750 mLx((0.65+0.35)/(0.65))=1200ml=1.2L...2 sig figs

5 0

3 years ago

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