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Kay [80]
2 years ago
11

The microscopes used today are just like the ones used by Leeuwenhoek and Hooke. True or false

Chemistry
1 answer:
Dmitry [639]2 years ago
6 0

Answer:

False

Explanation:

While we do know that A. Leeuwenhoek used a simple microscope that consisted of only 1 lens, Hooke used a compound microscope. Although, after trying a compound microscope, Hooke found out that it strained his eyes and continued to use a simple microscope for his <em>Micrographia</em>.

Thus, we can say that the (compound) microscopes used today are different than the (simple) microscope used by Hooke and Leeuwenhoek.

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Please help its urgent
mrs_skeptik [129]

Answer:

C

Explanation:

I picked C because the plant is interacting with the sun which is a nonliving part of the environment.

A, B and D are wrong because seaweed, horses, and trees are living things.

8 0
3 years ago
Read 2 more answers
Choose the FALSE statement.
Mariana [72]

Answer:

C. The half-life of C-14 is about 40,000 years.

Explanation:

The only false statement from the options is that the half-life of C-14 is 40,000yrs.

The half-life of an isotope is the time it takes for half of a radioactive material to decay to half of its original amount. C-14 has an half-life of 5730yrs. This implies that during every 5730yrs, C-14 will reduce to half of its initial amount.

  • All living organisms contain both stable C-12 and the unstable isotope of C-14
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5 0
3 years ago
How is enthalpy related to the spontaneity of a reaction?
mina [271]

Answer:

D

Explanation:

Gibb's free energy change(∆G) and Standard electrode potential of electrochemical (Ecell) determine the spontaneity of a reaction.

when ∆G > 0, the reaction is not spontaneous

∆G < 0, the reaction is spontaneous

∆G = 0, the reaction is in equilibrium

when Ecell > 0, the redox reaction is spontaneous

Ecell < 0, the redox reaction is not spontaneous

Ecell = 0, the redox reaction is in equilibrium.

5 0
3 years ago
What mass of natural gas (CH4) must you burn to emit 264 kJ of heat?
mariarad [96]
The heat released by cpmolete the combustion of organic products with oxygen is called heat of combustion.

Here 1 mol of CH4 realesed 802.3 KJ

To emit 264 kJ you multiply you need (1 mol of CH4/802.3 kJ)* 264 kJ = 0.329 mol of CH4

The molar mass, MM, of CH4 is 12 g/mol + 4*1g/mol = 16 g/mol

The to obtain the mass multiply the number of moles times the molar mass:

mass = n * MM = 0.329mol * 16g/mol = 5.26 grams

Answer: 5.26 grams



7 0
3 years ago
What data should be plotted to show that experimental concentration data fits a first-order reaction? g?
podryga [215]
I do not know i am doing this for the points
7 0
3 years ago
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