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3 years ago

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Pls help i will mark as brainliest

1 answer:

oee [108]3 years ago

8 0

Answer:

ano po ito?? isa po ba itong story

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lys-0071 [83]

The answer would be option B "I believe there is life on other planets." Scientific statements have a possibility to be wrong. It's not option A because option A is a opinion. It's not option C because option C is a fact. It's not option D because option D is a opinion.

Hope this helps!

8 0

3 years ago

Read 2 more answers

How many molecules are in 2.00 moles of H20????????

Elis [28]

2x 6.022x10^23= 1.204x10^24

7 0

3 years ago

What is the mass in grams of BaCl2 that is needed to prepare 200 mL of a 0.500 M solution

DENIUS [597]

Answer:

= 20.82 g of BaCl2

Explanation:

Given,

Volume = 200 mL

Molarity = 0.500 M

Therefore;

Moles = molarity × volume

= 0.2 L × 0.5 M

= 0.1 mole

But; molar mass of BaCl2 is 208.236 g/mole

Therefore; 0.1 mole of BaCl2 will be equivalent to;

= 208.236 g/mol x 0.1 mol

= 20.82 g

Therefore, the mass of BaCl2 in grams required is 20.82 g

6 0

3 years ago

Which molecule is amphoteric?

frosja888 [35]

Answer:

The answer is B

Explanation:

4 0

3 years ago

Read 2 more answers

A solid powder is known to be a mixture of NaCl and Na2CO3, but the relative amounts of each compound in the sample are unknown.

AveGali [126]

Answer:

Concentration of sodium carbonate in the solution before the addition of HCl is 0.004881 mol/L.

Explanation:

$Na_2CO_3(aq) +2 HCl(aq)\rightarrow 2NaCl(aq) + H_2O(l)+CO_2(g)$

Molarity of HCl solution = 0.1174 M

Volume of HCl solution = 83.15 mL = 0.08315 L

Moles of HCl = n

$molarity=\frac{moles}{Volume (L)}$

$0.1174 M=\frac{n}{0.08315 L}$

$n=0.1174 M\times 0.08315 L=0.009762 mol$

According to reaction , 2 moles of HCl reacts with 1 mole of sodium carbonate.

Then 0.009762 mol of HCl will recat with:

$\frac{1}{2}\times 0.009762 mol=0.004881 mol$

Moles of Sodium carbonate = 0.004881 mol

Volume of the sodium carbonate containing solution taken = 1L

Concentration of sodium carbonate in the solution before the addition of HCl:

$[Na_2CO_3]=\frac{0.004881 mol}{1 L}=0.004881 mol/L$

4 0

3 years ago