Government should use informed science to help make policies protect all citizens True or False

Pa help po science po

mamaluj [8]

Answer:

1. molecule, cell, tissue, organ, organ system, organism, population, community, ecosystem, biosphere

2. A

3. D

4. C

5. C

6. C

7. C

8. B

9. Sorry I don't know :(

10. B

11. A

12. C

13. Sorry :((

14. C

15. a

16. d

Sorry I don't know the rest.. I hope this helps though!!

4 0

1 year ago

2) 183 cg = kg 5 .00183 kg 0.0183 kg .183 kg 3) 0.25 kg = g

OlgaM077 [116]

Answer:

183 cg = 0.00183 kg

0.25 kg = 250 g

Explanation:

Use conversion factors. 1kg is equal to 1 x 10^5 cg (100000) and 1 kg is equal to 1 x 10^3 grams (1000 grams).

5 0

3 years ago

What type of fault usually occurs because of compression?

lianna [129]

Answer:c

Explanation:

4 0

2 years ago

Read 2 more answers

A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec

GalinKa [24]

Answer: The moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For lead (II) nitrate:

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol$

For NaI:

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol$

For the given chemical reaction:

$Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, $3.14\times 10^{-5}$ moles of NaI will react with = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol$ of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, $3.14\times 10^{-5}$ moles of NaI will produce = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles$ of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

4 0

2 years ago

Explain why 2-bromopropane reacts with sodium iodide in acetone over 104 times faster than bromocyclopropane. hint: examine the

Mama L [17]

Answer:-

The reaction of 2-bromopropane reacts with sodium iodide in acetone is an example of Sn2 reaction.

The I - attacks from backside to give the transition state for both.

If we compare the transition state for cyclobromopropane 2-bromopropane then we see in case of cyclobromopropane transition state, one of the H is very close to the incoming I -.

This results in steric strain and less stability of the transition state. Hence 2-bromopropane reacts with sodium iodide in acetone over 104 times faster than bromocyclopropane.

7 0

3 years ago