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Elenna [48]
3 years ago
5

Jack hypothesize that more mold will grow on a slice of bread that is left in the dark that a slice of bread left in the light h

e wants to do in experiment to test this hypothesis Jack has four slices of bread for this experiment was condition should not be the same for the four slices of bread in order to test a hypothesis a The type of bread b used to be where the bread is placed c how long the bread is being observed The amount of jelly on them
Chemistry
1 answer:
cestrela7 [59]3 years ago
7 0
A- The type of bread. some breads can become moldy faster than others.
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Are these answers correct? I’m a little unsure
Naily [24]

Answer:

yes I think that they are correct

8 0
3 years ago
The Russian chemist who designed the periodic table that was similar to the Periodic Table used today was _______________-
Xelga [282]
Dmitri Mendeleev, he realized that the physical and chemical properties of elements were related to their atomic mass and arranged them into groups (columns)
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4 years ago
A 3" diameter germanium wafer that is 0.020" thick at 300K has 3.419*10^17 As atoms added to it. What is the resistivity of the
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3 years ago
What mass of butane in grams is necessary to produce 1.5×103 kJ1.5×103 kJ of heat? What mass of CO2CO2 is produced? Assume the r
saul85 [17]

32.8 g of Butane is required and 99.3 g of CO₂ is produced

<u>Explanation:</u>

The above mentioned reaction can be written as,

C₄H₁₀(g) + 13 O₂(g) → 4CO₂(g) + 5 H₂O(g)     where ΔH (rxn)= -2658 kJ

It is given that 1.5 × 10³ kJ of energy is produced, the original reaction says that 2658 kJ of heat is produced, which means that less than one mole of butane is used in the reaction.

That is

$\frac{1500}{2658}=0.564 \text { moles }    of butane reacted

Now this moles is converted into mass by multiplying it with its molar mass  = 0.564 mol × 58.122 g / mol

                     = 32.8 g of butane.

Mass of CO₂ produced = 0.564 ×44.01 g /mol × 4 mol

                                        = 99.3 g of CO₂

Thus 32.8 g of Butane is required and 99.3 g of CO₂ is produced

7 0
3 years ago
How many L of a 7.5 H2SO4 stock solution would you need to prepare a dilute solution 100 L of 0.25M H2SO4
blsea [12.9K]

Answer:

3.33 L

Explanation:

We can solve this problem by using the equation:

  • C₁V₁=C₂V₂

Where the subscript 1 refers to one solution and subscript 2 to the another solution, meaning that in this case:

  • C₁ = 0.25 M
  • V₁ = 100 L
  • C₂ = 7.5 M
  • V₂ = ?

We input the data:

  • 0.25 M * 100 L = 7.5 M * V₂
  • V₂ = 3.33 L

Thus the answer is 3.33 liters.

3 0
3 years ago
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