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3 years ago

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

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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

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3 years ago

(select all that apply.) the elements most often found in organic molecules are _____.

Arturiano [62]

C. Carbon. H. Hydrogen. N. Nitrogen. O. Oxygen. P. Phosphorus. S. Sulfur.

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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

519.01 m/s

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

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

$T=345.60K$

Now , Use Vrms to calculate the velocity of O2

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

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

$v_{rms}=\sqrt{26935.001}$

$v_{rms}=519.01m/s$

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