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3 years ago

Which is not a standard type of phytoplankton

Chemistry

1 answer:

GalinKa [24]3 years ago

8 0

I think it is answer choice A:coral im not rlly sure but im kinda thinking choice A

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Imagine that you have 100g of water.

laiz [17]

Answer:

Explanation:

So, just find 55 on the solubility side and with your finger just move to the right untill you touch the line you should get in between 46-48. I would go more for 48.

Hope you have a wonderful day!

6 0

3 years ago

For the balanced equation shown below, how many moles of Al will react with 98.7 moles of CuSO4

xz_007 [3.2K]

Answer:

65.8 mol

Explanation:

Step 1: Write the balanced equation for the single displacement reaction

2 Al + 3 CuSO₄ → Al₂(SO₄)₃ + 3 Cu

Step 2: Establish the appropriate molar ratio

According to the balanced equation, the molar ratio of Al to CuSO₄ is 2:3.

Step 3: Calculate the number of moles of Al that will react with 98.7 moles of CuSO₄

98.7 mol CuSO₄ × 2 mol Al/3 mol CuSO₄ = 65.8 mol Al

6 0

3 years ago

This infected cut has activated an ________ response<br><br> inflammatory<br> O pathogens<br> O skin

qwelly [4]

Answer:

Inflammatory

Explanation:

5 0

3 years ago

How many grams of solute are needed to order to prepare 100.00 mL of a 0.1000 M solution of a compound with a molecular weight o

Troyanec [42]

The grams of solute are required.

The mass of solute is 3.5 g

c = Molarity = 0.1 M

M = Molar mass = 350 g/mol

V = Volume of solution = 100 mL = 0.1 L

n = Number of moles

m = Mass of solute

Molarity is given by

$c=\dfrac{n}{V}\\\Rightarrow n=cV\\\Rightarrow n=0.1\times 0.1\\\Rightarrow n=0.01\ \text{moles}$

Molar mass is given by

$M=\dfrac{m}{n}\\\Rightarrow m=Mn\\\Rightarrow m=350\times 0.01\\\Rightarrow m=3.5\ \text{g}$

The mass of solute is 3.5 g

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7 0

2 years ago

Read 2 more answers

At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react

erastova [34]

Answer:

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

$N_{2 (g)}$ + $O_{2 (g)}$ ⇄ 2 $NO_{(g)}$

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no $N_{2 (g)}$ nor $O_{2 (g)}$ present. Immediately, $N_{2 (g)}$ and $O_{2 (g)}$ are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [ $N_{2 (g)}$ ]=0 ; [ $O_{2 (g)}$ ]= 0 ; [ $NO_{(g)}$ ]=0.60M

C: [ $N_{2 (g)}$ ]=+x ; [ $O_{2 (g)}$ ]= +x ; [ $NO_{(g)}$ ]=-2x

E: [ $N_{2 (g)}$ ]=0+x ; [ $O_{2 (g)}$ ]= 0+x ; [ $NO_{(g)}$ ]=0.60-2x

Now we can use the constant information:

$K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }$