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notka56 [123]
3 years ago
7

What observations and inferences can you make from this picture?

Chemistry
2 answers:
Salsk061 [2.6K]3 years ago
6 0
I know they are fishing and there is an alligator in the lake. The alligator might grab the bait and attack the fishers.

Hope I helped!
Let me know if you need anything else!
~ Zoe
Liono4ka [1.6K]3 years ago
4 0

Answer:

Sample Response: The people are part of a softball team. They appear happy, they are showing the number one with their fingers, and they are holding a trophy. They probably won the softball league championship.

Explanation:

e2020

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When you decrease the volume of a gas, what happens to the temperature?
ale4655 [162]

the temperature is decreased

4 0
3 years ago
Calculate the cost of gasoline for a 320 mile trip from Boston to New
Talja [164]
The cost of gasoline for the trip would cost $57.28 because you first divide 320 by 20 which gives you 16 than after that you do 16 multiplied by 3.58 which gives you $57.28 as your final product
5 0
3 years ago
(select all that apply.) the elements most often found in organic molecules are _____.
Arturiano [62]
<span>C. Carbon. H. Hydrogen. N. Nitrogen. O. Oxygen. P. Phosphorus. <span>S. Sulfur.</span></span>
7 0
3 years ago
Read 2 more answers
If He gas has an average kinetic energy of 4310 J/mol under certain conditions, what is the root mean square speed of O2 gas mol
Free_Kalibri [48]

Answer:

The root mean square speed of O2 gas molecules is

<u>519.01 m/s</u>

<u></u>

Explanation:

The root mean square velocity  :

v_{rms}=\sqrt{\frac{3RT}{M}}

K.E_{avg}=\frac{3}{2}RT

K.E =\frac{1}{2}mv_{rms}^{2}

Molar mass , M

For He = 4 g/mol

For O2 = 2 x 16 = 32 g/mol

O2 = 32/1000 = 0.032 Kg/mol

First calculate the temperature at which the K.E of He is 4310J/mol

K.E of He =

K.E_{avg}=\frac{3}{2}RT

T=\frac{2(K.E)}{3(R)}

K.E of He = 4310 J/mol

T=\frac{2(4310J/mol)}{3(8.314J/Kmol)}

T=345.60K

<u>Now , Use Vrms to calculate the velocity of O2</u>

v_{rms}=\sqrt{\frac{3(8.314J/Kmol)(345.60K)}{0.032Kg/mol}}

v_{rms}=\sqrt{\frac{8619.9552}{0.032}}

v_{rms}=\sqrt{26935.001}

v_{rms}=519.01m/s

6 0
3 years ago
If 6.00 g of CaCl2 • 2 H2O and 5.50 g of Na2CO3 are allowed to react in aqueous solution, what mass of CaCO3 will be produced? P
Andre45 [30]

Answer:

6.00 g CaCl₂ .2H₂O /1 × 1 mol CaCl₂ .2H₂O / 147 g CaCl₂ .2H₂O × 0.04 mol CaCO₃/ 0.04 mol of  CaCl₂ .2H₂O  × 100 g CaCO₃ /  1 mole CaCO₃ = 4 g

5.50 g Na₂CO₃   /1 × 1 Na₂CO₃  / 106 g Na₂CO₃ × 0.05 mol CaCO₃/ 0.05 mol of Na₂CO₃  × 100 g CaCO₃ /  1 mole CaCO₃ = 5 g

Explanation:

Given data:

Mass of CaCl₂.2H₂O = 6.00 g

Mass of Na₂CO₃ = 5.50 g

Mass of CaCO₃ produced = ?

Solution:

Number of moles of CaCl₂.2H₂O.

Number of moles = mass/ molar mass

Number of moles = 6.00 g/ 147 g/ mol

Number of moles = 0.04 mol

Number of moles of Na₂CO₃:

Number of moles = mass/ molar mass

Number of moles = 5.50 g/ 106 g/ mol

Number of moles = 0.05 mol

Chemical equation:

CaCl₂  +  Na₂CO₃   →   CaCO₃ + 2NaCl

Now we will compare the moles of CaCO₃  with Na₂CO₃  and CaCl₂ through balanced chemical equation .

                      CaCl₂              :               CaCO₃

                             1                :                1

                       0.04               :            0.04

Mass of CaCO₃:

Mass = number of moles × molar mass

Mass = 0.04 mol× 100 g/mol

Mass = 4 g

6.00 g CaCl₂ .2H₂O /1 × 1 mol CaCl₂ .2H₂O / 147 g CaCl₂ .2H₂O × 0.04 mol CaCO₃/ 0.04 mol of  CaCl₂ .2H₂O  × 100 g CaCO₃ /  1 mole CaCO₃ = 4 g

                     Na₂CO₃            :            CaCO₃

                          1                   :                1

                       0.05               :            0.05

Mass of CaCO₃:

Mass = number of moles × molar mass

Mass = 0.05 mol× 100 g/mol

Mass = 5 g

5.50 g Na₂CO₃   /1 × 1 Na₂CO₃  / 106 g Na₂CO₃ × 0.05 mol CaCO₃/ 0.05 mol of Na₂CO₃  × 100 g CaCO₃ /  1 mole CaCO₃ = 5 g

                     

5 0
3 years ago
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