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loris [4]
3 years ago
11

Gold ions occur in seawater to the extent of1.0.

Chemistry
1 answer:
faltersainse [42]3 years ago
5 0

Answer:

According to research, the concentration of gold ions in the seawater is 10⁻¹⁴g/cm³.

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the chemical compound C2F4 is used to make PTEE (Teflon). How manyC2F4 molecules are in 485 kilograms of this material?
monitta

Answer:

molecules=2.92x10^{27}moleculesC_2F_4

Explanation:

Hello,

In this case, we use the Avogadro's number to compute the molecules of C2F4 whose molar mass is 100 g/mol contained in a 485-kg sample as shown below:

molecules=485kgC_2F_4*\frac{1000gC_2F_4}{1kgC_2F_4} *\frac{1molC_2F_4}{100gC_2F_4}*\frac{6.022x10^{23}molecules C_2F_4}{1molC_2F_4}  \\\\molecules=2.92x10^{27}moleculesC_2F_4

Best regards,

5 0
3 years ago
Giving brainliest and thanks to best answer <333 :>
valkas [14]

Answer:

C

Explanation:

radiation from the sun first warms the outer atmosphere (trophosphere)

convection(aka just heat moving through gas or liquid) brings the warmth down lower

conduction heats the ground

gl lol :))

7 0
3 years ago
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I’m a paragraph of no less than 125 words describe the steps you can take to effectively organize your writing
Tanya [424]

how many sentences must it be?

5 0
3 years ago
If 50 ml of 0.235 M NaCl solution is diluted to 200.0 ml what is the concentration of the diluted solution
Helen [10]

This is a straightforward dilution calculation that can be done using the equation

M_1V_1=M_2V_2

where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.

Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:

M_2=\frac{M_1V_1}{V_2}.

Substituting in our values, we get

\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].

So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.

5 0
3 years ago
Please do this please please it depends on my grades
solong [7]

Answer:

I think it's

Explanation:

It's A and C

I guess...

Sry if I'm wrong

4 0
2 years ago
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