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garik1379 [7]
3 years ago
14

What did early scientists assume that the polar caps had in common?

Chemistry
1 answer:
Vilka [71]3 years ago
3 0

Answer:

They assumed they both had water.

Explanation:

Because they only could look at it through telescopes that were not advanced

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The study of organisms in relationship to their environment is .
laiz [17]

Answer:

Ecology.

Explanation:

Because you are epic.

4 0
3 years ago
Read 2 more answers
a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re
Blababa [14]

Answer: 41.5 mL

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n}{V_s}

where,

n = moles of solute

V_s = volume of solution in L

Given : 59.4 g of H_2SO_4 in 100 g of solution  

moles of H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61

Volume of solution =\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml

Now put all the given values in the formula of molality, we get

Molality=\frac{0.61\times 1000}{54.6ml}=11.2M

To calculate the volume of acid, we use the equation given by neutralisation reaction:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of stock acid which is H_2SO_4

M_2\text{ and }V_2 are the molarity and volume of dilute acid which is H_2SO_4

We are given:

M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL

Putting values in above equation, we get:

11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

4 0
3 years ago
How many moles of oxygen are needed to react with 23.8 moles of aluminum?
antoniya [11.8K]

Answer:

17.85moles

Explanation:

Check attachment

7 0
3 years ago
How to do my homework? i’ll give 20 points to anyone who will do it!
tensa zangetsu [6.8K]

what is it ??? ill edit this

3 0
3 years ago
Read 2 more answers
Balance this equation. Pb(NO3)2(aq)+NaCl(aq) -> NaNO3(aq)+PbCl2(s)
const2013 [10]

Answer:

Pb(NO3)2(aq) + 2NaCl(aq) -> 2NaNO3(aq)+PbCl2(s)

Explanation:

Pb(NO3)2(aq)+NaCl(aq) -> NaNO3(aq)+PbCl2(s)

This is how it starts out.

Left:

  • 2 NO3s
  • 1 Pb
  • 1 Na
  • 1 Cl

Right

  • 1 Na
  • 1 NO3
  • 1 Pb
  • 2 Cl

So the place to start with this equation is to bring the Cls up to 2

Pb(NO3)2(aq)+2NaCl(aq) -> NaNO3(aq)+PbCl2(s)

But the Nas are now out of kilter.

Pb(NO3)2(aq)+ 2NaCl(aq) -> NaNO3(aq)+PbCl2(s)

Now the right has a problem. There's only 1 Na

Pb(NO3)2(aq) + 2 NaCl(aq) -> 2NaNO3(aq)+PbCl2(s)

Check it out. It looks like we are done.

5 0
2 years ago
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