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3 years ago

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3. Sketch a simple model that shows why the constellations change over the course of a year. Then, annotate the model to explain

why you see different stars throughout the year.

1 answer:

Lorico [155]3 years ago

8 0

Constellation and i will not be in a new

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Given 50.0 grams of Strontium, find grams of Phosphorus required for complete reaction

Sveta_85 [38]

Answer:

Mass = 11.78 g of P₄

Explanation:

The balance chemical equation is as follow:

6 Sr + P4 → 2 Sr₃P₂

Step 1: Calculate moles of Sr as;

Moles = Mass / M/Mass

Moles = 50.0 g / 87.62 g/mol

Moles = 0.570 moles

Step 2: Find moles of P₄ as;

According to equation,

6 moles of Sr reacted with = 1 mole of P₄

So,

0.570 moles of Sr will react with = X moles of P₄

Solving for X,

X = 1 mol × 0.570 mol / 6 mol

X = 0.0952 mol of P₄

Step 3: Calculate mass of P₄ as,

Mass = Moles × M.Mass

Mass = 0.0952 mol × 123.89 g/mol

Mass = 11.78 g of P₄

8 0

3 years ago

Identify the correct name for each compound.<br> NaOH:<br> CaSO4<br> NHACN:<br> Al2(SO4)3

klemol [59]

Answer:

Sodium hydroxide

Calcium sulfate

Ammonium cyanide

Aluminum sulfate

Explanation:

Just did it

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3 years ago

If 12.4 mol of Ne gas occupies 122.8 L, how many mol of Ne would occupy 339.2 L under the same temperature and pressure? Record

8090 [49]

Answer:

3.43×10¹ mol

Explanation:

Given data:

Initial number of moles = 12.4 mol

Initial volume = 122.8 L

Final number of moles = ?

Final volume = 339.2 L

Solution:

The number of moles and volume are directly proportional to each other at same temperature and pressure.

V₁/n₁ = V₂/n₂

122.8 L/ 12.4 mol = 339.2 L / n₂

n₂ = 339.2 L× 12.4 mol / 122.8 L

n₂ = 4206.08 L.mol /122.8 L

n₂ = 34.3mol

In scientific notation:

3.43×10¹ mol

7 0

3 years ago

What are the 10 element

maw [93]

hydrogen
oxygen
carbon
nitrogen
fluorine
silicone
boron
argon
cobalt
aluminum

hope it hepls✨

4 0

2 years ago

How many Joules are released to cool 250.0 grams of liquid water from 100°C to 0°C? The specific heat of water is 4.180 J/g.C.

nlexa [21]

$\bold{\huge{\orange{\underline{ Solution}}}}$

$\bold{\underline{ Given :- }}$

<u>We </u><u>have </u><u>250g </u><u>of </u><u>liquid </u><u>water </u><u>and </u><u>it </u><u>needs </u><u>to </u><u>be </u><u>cool </u><u>at </u><u>temperature </u><u>from </u><u>1</u><u>0</u><u>0</u><u>°</u><u> </u><u>C </u><u>to </u><u>0</u><u>°</u><u> </u><u>C</u>
<u>Specific </u><u>heat </u><u>of </u><u>water </u><u>is </u><u>4</u><u>.</u><u>1</u><u>8</u><u>0</u><u>J</u><u>/</u><u>g</u><u>°</u><u>C</u>

$\bold{\underline{ To \: Find :- }}$

<u>We </u><u>have </u><u>to </u><u>find </u><u>the</u><u> </u><u>total</u><u> </u><u>number </u><u>of </u><u>joules </u><u>released</u><u>. </u>

$\bold{\underline{ Let's \:Begin:- }}$

<u>We </u><u>know </u><u>that</u><u>, </u>

Amount of heat energy = mass * specific heat * change in temperature

<u>That </u><u>is, </u>

$\sf{\red{ Q = mcΔT }}$

<u>Subsitute </u><u>the </u><u>required </u><u>values </u><u>in </u><u>the </u><u>above </u><u>formula </u><u>:</u><u>-</u>

$\sf{ Q = 250 × 4.180 ×(0 - 100 )}$

$\sf{ Q = 250 × 4.180 × - 100 }$

$\sf{ Q = 250 × - 418}$

$\sf{\pink{ Q = - 104,500 J }}$

Hence, 104,500 J of heat is released to cool 250 grams of liquid water from 100° C to 0° C.

$\bold{\underline{ Now :- }}$

<u>We </u><u>have </u><u>to </u><u>tell </u><u>whether </u><u>the </u><u>above </u><u>process </u><u>is </u><u>endothermic </u><u>or </u><u>exothermic </u><u>:</u><u>-</u>

Here, In the above process ΔT is negative and as a result of it Q is also negative that means above process is Exothermic

<u>Exothermic </u><u>process </u><u>:</u><u>-</u><u> </u><u>It </u><u>is </u><u>the </u><u>process </u><u>in </u><u>which </u><u>heat </u><u>is </u><u>evolved </u><u>. </u>
<u>Endothermic </u><u>process </u><u>:</u><u>-</u><u> </u><u>It </u><u>is </u><u>the </u><u>process </u><u>in </u><u>which </u><u>heat </u><u>is </u><u>absorbed </u><u>.</u>

4 0

2 years ago