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Natasha_Volkova [10]
3 years ago
13

1. Complete the following Data Table as you conduct the lab

Chemistry
2 answers:
miskamm [114]3 years ago
4 0

Answer:

Do to half of the mnairals this can not be made into a lab there is an error

Explanation:

Alona [7]3 years ago
4 0

Answer:

Do to half of the mnairals this can not be made into a lab there is an error

Explanation:

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True or False? In a C++ floating-point constant, a decimal point is not required if exponential (E) notation is used.
defon

Answer:

The answer is True

Explanation:

This is because they are equivalent and have same number of bit.

7 0
3 years ago
What population would be hurt if the bear population was to increase
KengaRu [80]
Fish would be hurt because that’s what bears mostly eat
7 0
3 years ago
How many formula units are in 4.52 moles of H3SO3?<br> Type your answer
lakkis [162]

Answer:

The answer is 98.07848. We assume you are converting between grams H2SO4 and mole. You can view more details on each measurement unit: This compound is also known as Sulfuric Acid. The SI base unit for amount of substance is the mole. 1 grams H2SO4 is equal to 0.010195916576195 mole.

<u>Quick conversion chart of moles H2SO3 to grams</u>

1 moles H2SO3 to grams = 82.07908 grams

2 moles H2SO3 to grams = 164.15816 grams

3 moles H2SO3 to grams = 246.23724 grams

4 moles H2SO3 to grams = 328.31632 grams

5 moles H2SO3 to grams = 410.3954 grams

6 moles H2SO3 to grams = 492.47448 grams

7 moles H2SO3 to grams = 574.55356 grams

8 moles H2SO3 to grams = 656.63264 grams

9 moles H2SO3 to grams = 738.71172 grams

10 moles H2SO3 to grams = 820.7908 grams

7 0
3 years ago
When 45 g of an alloy, at 25oC, are dropped into 100.0 g of water, the alloy absorbs 956 J of heat. If the final temperature of
taurus [48]

Answer:

The answer to the question is

The specific heat capacity of the alloy = 1.77 J/(g·°C)

Explanation:

To solve this, we list out the given variables thus

Mass of alloy = 45 g

Initial temperature of the alloy = 25 °C

Final temperature of the alloy = 37 °C

Heat absorbed by the alloy = 956 J

Thus we have

ΔH = m·c·(T₂ - T₁) where  ΔH = heat absorbed by the alloy = 956 J, c = specific heat capacity of the alloy and T₁ = Initial temperature of the alloy = 25 °C , T₂ = Final temperature of the alloy = 37 °C  and m = mass of the alloy = 45 g

∴ 956 J = 45 × C × (37 - 25) = 540 g·°C×c  or

c = 956 J/(540 g·°C) = 1.77 J/(g·°C)

The specific heat capacity of the alloy is 1.77 J/(g·°C)

3 0
3 years ago
A mixture of chalk powder and water can be
grin007 [14]

Answer:

<h3>Right answer is: ( a) chalk powder remains suspended in water.</h3>

Explanation:

Filtration is the technique used to separate suspended solute particles from a solution . The chalk powder remains suspended in the solution and can easily be filtered through a filter paper , the chalk powder can be collected on the filter paper and clear solvent is collected as the filtrate.

4 0
3 years ago
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