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hichkok12 [17]
2 years ago
13

True or false? Second level consumers may be carnivore or omnivores

Chemistry
1 answer:
const2013 [10]2 years ago
3 0
Yes this is true because they are able to consume 2 organisms which are plants and animal just like us.
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A student did an experiment to determine the specific heat capacity of a metal alloy. The student put a sample of the alloy in b
guapka [62]

Answer:

a) ∆T=T1-T2

b) At the particle level the temperature changes are the result of the added energy causing the particles of water to move more vigorously. Either the particles of solid vibrate more vigorously about their fixed positions or the particles of liquid and gas move about their container more rapidly.

c) The state in which two substances in physical contact do not share any heat energy. The temperature of two substances in thermal equilibrium is said to be the same. Also see thermodynamics.

Explanation:

hope that helped good luck!

7 0
2 years ago
How can you increase the momentum of an object?
Elena-2011 [213]
I think its B because if u increase the mass itll have more force which will increase the momentum
5 0
3 years ago
Besides water, which of the following compounds are found most in sea water?
Tju [1.3M]
The answer is sodium chloride.

Explanation:

Sodium chloride refers to table salt, and is the most abundant of salts found in seawater.
4 0
2 years ago
g A 0.4395 g sample of aluminum reacts according to our experiment to produce alum. 5.1629 g of dried alum crystals are recovere
Alex

Answer:

92.75%

Explanation:

The overall chemical equation for the reaction in the preparation of alum from the aluminium can be expressed as:

\mathtt{2Al + 2KOH + 4H_2SO_4 +2H_2O   \to 2KAl(SO_4)_2 2H_2O +3H_2}

From above; we will see that 2 moles of Aluminium react with sulphuric acid and water to produce 2 moles o aluminium alum.

Therefore, the theoretical yield can be determined as:

=0.4395 \ g Al \times \dfrac{1 \ mol \ Al}{27 \ g Al }\times \dfrac{2 \ mol \ KAl(SO_4)_2}{2 \  mol \ Al}\times \dfrac{294.23 \ KAl(SO_4)_2}{1 \ mol \ KAl(SO_4)_2}

= 4.789g of KAl(SO_4)_2

To find the percent yield, we need to divide the actual yield by the theoretical yield and then multiply it with 100.

∴

percent yield = ( mass of alum(g)/theoretical yield(g) ) × 100

percent yield = ( 4.789g / 5.1629g ) × 100%

percent yield = 0.9275 × 100%

percent yield = 92.75%

Thus, the percent yield of the experiment 92.75%

8 0
3 years ago
Convert a pressure of 1.75 atm to kPa and to mm Hg
Serjik [45]

P = 1.75 atm = __ kPa =

_mmHg

1 atm = 760 mm Hg

We can write the two conversion factors

1 atm / 760 mm Hg or 760 mm Hg / 1 atm

1.75 atm x ( 760 mm Hg \ 1 atm)

1330 mm Hg

1 atm = 101325 pa

We can write two conversion factors

1 atm / 101325 pa or 101325 pa / 1 atm

1.75 atm x ( 101325 pa / 1 atm )

177319 pa or 177.319 kPa

3 0
2 years ago
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