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Kitty [74]
2 years ago
9

I need help fast!! Djdndndndndndnnds

Chemistry
1 answer:
adell [148]2 years ago
6 0

Answer:

1st one: carbon dioxide

2nd: Glucose

3rd: Oxygen

4th: Energy

Sorry if this is wrong because I don't see the options!

Hope this helps!

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The labeled images each represent the wave patterns found in the electromagnetic wave spectrum. Which image is correctly labeled
Nataly_w [17]
Https://us-static.z-dn.net/files/d15/c111c5e1b23135c61adec7b554629964.jpg

I believe the answer is the third option. Hope this helps! Please tell me if I am wrong or if there was an error in my answer... also sorry this answer is late.

6 0
3 years ago
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The reaction na 3po 4( aq) 3 agno3( aq) → ag 3po 4( s) 3 nano 3( aq) is best classified as a(n):_____.
Ugo [173]

The reaction  Na_{3} PO_{4} ( aq) +3AgNO_{3} (aq) → Ag_{3} PO_{4} ( aq) +3NaNO_{3} (aq) is best classified as double displacement reaction.

Those reaction in which two compounds react by exchanges of ions to form two new compounds is called  double displacement reaction. The easiest way to identify double displacement reactions is to check to see whether the cations exchanged anions with each other  or not . Always balanced chemical equation is used to determine.

There are three types of double displacement reaction which is given as,

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The real world example of double displacement reaction is combining vinegar and baking soda to create homemade volcano.

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6 0
1 year ago
What is significant about 42?​
maks197457 [2]

Answer:

It’s twice as much as 9+10

Explanation:

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2 years ago
How much energy does it take to boil 100 mL of water?
yawa3891 [41]

Answer:

the answer is D. Because the 1 atm pressure of water is 40.65 or 40.7.

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An atom of gold has a mass of 3.271 X 10-22 g. How many atoms of gold are in 5.00 g of gold? (Give your answer in scientific not
Kobotan [32]

Answer:

1.53 × 10²² atoms Ag

Explanation:

Step 1: Define conversions

3.271 × 10⁻²² g = 1 atom

Step 2: Use Dimensional Analysis

5.00 \hspace{3} g \hspace{3} Ag(\frac{1 \hspace{3} atom \hspace{3} Ag}{3.271(10)^{-22} \hspace{3} g \hspace{3} Ag} ) = 1.52858 × 10²² atoms Ag

Step 3: Simplify

We have 3 sig figs.

1.52858 × 10²² atoms Ag ≈ 1.53 × 10²² atoms Ag

5 0
2 years ago
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