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ddd [48]
3 years ago
5

Helppp nowwww plssssss!!

Chemistry
1 answer:
Anna007 [38]3 years ago
4 0
PLS HELP ME ASAP I DONT HAVE TIME. IT ALSO DETECTS IF ITs RIGHT OR WRONG. PLS HELP ME ASAP I DONT HAVE TIME. IT ALSO DETECTS IF ITs RIGHT OR WRONG. PLS HELP ME ASAP I DONT HAVE TIME. IT ALSO DETECTS IF ITs RIGHT OR WRONG. PLS HELP ME ASAP I DONT HAVE TIME. IT ALSO DETECTS IF ITs RIGHT OR WRONG. PLS HELP ME ASAP I DONT HAVE TIME. IT ALSO DETECTS IF ITs RIGHT OR WRONG. PLS HELP ME ASAP I DONT HAVE TIME. IT ALSO DETECTS IF ITs RIGHT OR WRONG. PLS HELP ME ASAP I DONT HAVE TIME. IT ALSO DETECTS IF ITs RIGHT OR WRONG. PLS HELP ME ASAP I DONT HAVE TIME. IT ALSO DETECTS IF ITs RIGHT OR WRONG.
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Cell,neclus,genes,chromosomes,tissue,organ,organ system organism arrange in order?
egoroff_w [7]
The order in increasing is:

genes, chromosomes, cells, tissue, organ, system, organism
7 0
3 years ago
Write the balanced equation. Then calculate the volume of 0.65 M HCl required to completely neutralize 400.0 ml of 0.88 M KOH.
Misha Larkins [42]
Hello!

The balanced equation for the neutralization of KOH is the following:

HCl(aq) + KOH(aq) → KCl(aq) + H₂O(l)

To calculate the volume of HCl required, we can apply the following equation:

 moles HCl = moles KOH \\  \\ cHCl*vHCl=cKOH*vKOH \\  \\ vHCl= \frac{cKOH*vKOH}{cHCl}= \frac{400 mL*0,88M}{0,65M}=  541,54mL

So, the required volume of HCl is 541,54 mL

Have a nice day!
8 0
3 years ago
Convert 150 g/L to the unit g/mL.<br><br> 15,000 g/mL<br> 15 g/mL<br> 0.15 g/mL<br> 0.0015 g/mL
fredd [130]
1 g/L ------- 0.001 g/mL
150 g/L ----- ?

150 x 0.001 / 1

= 0.15 g/mL

Answer C
6 0
4 years ago
Read 2 more answers
The Pecos Pueblo developed as a trading center over time. What evidence in the text supports the idea that this took a long time
Ymorist [56]

Answer:

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Explanation:

3 0
3 years ago
What is the mass (in grams) of 9.93 Ă— 1024 molecules of methanol (CH3OH)?
WINSTONCH [101]

Answer is: mass of methanol is 528.32 grams.

N(CH₃OH) = 9.93·10²⁴; number of methanol molecules.

n(CH₃OH) = N(CH₃OH) ÷ Na (Avogadro constant).

n(CH₃OH) = 9.93·10²⁴ ÷ 6.022·10²³ 1/mol.

n(CH₃OH) = 16.49 mol; amount of substance.

m(CH₃OH) = n(CH₃OH) · M(CH₃OH).

m(CH₃OH) = 16.49 mol · 32.04 g/mol.

m(CH₃OH) = 528.32 g.

7 0
3 years ago
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