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3 years ago

Which of the following questions is most testable through scientific experimentation?

2 answers:

3 0

The Question that I believe that is the most testable through scientific experimentation would be the first question.

Which gelatin takes longer to set. Gelatin with fruit or gelatin without fruit.

Genrish500 [490]3 years ago

3 0

Answer:

Which gelatin takes longer to set: gelatin with fruit or gelatin without fruit?

Explanation:

Option A is Correct. In chemistry the rate of reaction or kinetics is always studied so this is most testable through scientific experimentation

Option B is Incorrect. This is more relevant in the hospitality business and is not testable through scientific experimentation.

Option C is Incorrect. This is more relevant in the hospitality business.

Option D is Incorrect. This has more to do with commerce or economics.

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Gabriel and his friend found a piece of metal which they want to

Dominik [7]

50/5.2 that’s the equation that you have to solve then whatever comes out yo calculator is the answer

6 0

3 years ago

Read 2 more answers

The Nutrition Facts label for crackers states that one serving contains 19 g of carbohydrate, 4 g of fat, and 2 g of protein. Wh

sp2606 [1]

Answer: The amount of kilocalories contained in the given serving of crackes is 0.120 kCal

Explanation:

As per the USDA:

Carbohydrates provide 4 calories per gram, protein provides 4 calories per gram, and fat provides 9 calories per gram

We are given:

Mass of fat in the crackers = 4 g

Mass of carbohydrates in the crackers = 19 g

Mass of protein in the crackers = 2 g

Conversion factor used: 1 kCal = 1000 Cal

Applying unitary method:

For Fat:

1 gram of fat provides 9 calories

So, 4 gram of fat will provide = $\frac{9}{1}\times 4=36Cal=0.036kCal$

For Carbohydrates:

1 gram of carbohydrates provides 4 calories

So, 19 gram of carbohydrates will provide = $\frac{4}{1}\times 19=76Cal=0.076kCal$

For Proteins:

1 gram of proteins provides 4 calories

So, 2 gram of fat will provide = $\frac{4}{1}\times 2=8Cal=0.008kCal$

Total kilocalories per serving = [0.036 + 0.076 + 0.008] kCal = 0.120 kCal

Hence, the amount of kilocalories contained in the given serving of crackes is 0.120 kCal

8 0

3 years ago

Pls someone help me out

SpyIntel [72]

Do you still need help

4 0

2 years ago

A sealed reaction vessel initially contains 1.113×10-2 moles of water vapor and 1.490×10-2 moles of CO(g).

Molodets [167]

Answer:

Kc = $\frac{[8.326x10-3]^{1} }{[1.113x10-2]^{1}[1.490x10-2]^{1} }$

Kc = 50.2059

Explanation:

1. Balance the equation

2. Use the Kc formula

Remember that pure substances, like H2 are not included on the Kc formula

4 0

3 years ago

What is the molality of a solution if 100.0 g of glucose (C&Hi20e) were dissolved into 750. mL of water?

hodyreva [135]

Answer: The molality of solution is 0.740 m.

Explanation:

To calculate the mass of solvent (water), we use the equation:

$Density=\frac{Mass}{Volume}$

Volume of water = 750 mL

Density of water = 1 g/mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{750mL}\\\\\text{Mass of water}=750g$

To calculate the molality of solution, we use the equation:

$Molarity=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(C_6H_{12}O_6)$ = 100.0 g

$M_{solute}$ = Molar mass of solute $(C_6H_{12}O_6)$ = 180 g/mol

$W_{solvent}$ = Mass of solvent (water) = 750 g

Putting values in above equation, we get:

$\text{Molality of }C_6H_{12}O_6=\frac{100\times 1000}{180\times 750}\\\\\text{Molality of }C_6H_{12}O_6=0.740m$

Hence, the molality of solution is 0.740 m.

6 0

3 years ago