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3 years ago

What contains most of earths water

Chemistry

1 answer:

nikitadnepr [17]3 years ago

5 0

The ocean hold 97 percent of earths water.

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Starting from a 1 M stock of NaCl, how will you make 50 ml of 0.15 M NaCl?

WARRIOR [948]

Answer:

We need 7.5 mL of the 1M stock of NaCl

Explanation:

Data given:

Stock = 1M this means 1 mol/ L

A 0.15 M solution of 50 mL has 0.0075 moles NaCl per 50 mL

Step 2: Calculate the volume of stock we need

The moles of solute will be constant

and n = M*V

M1*V1 = M2*V2

⇒ with M1 = the initial molair concentration = 1M

⇒ with V1 = the volume we need of the stock

⇒ with V2 = the volume we want to make of the new solution = 50 mL = 0.05 L

⇒ with M2 = the concentration of the new solution = 0.15 M

1*V1 = 0.15*(50)

V1 = 7.5 mL

Since 0.0075 L of 1M solution contains 0.0075 moles

50 mL solution will contain also 0.0075 moles but will have a molair concentration of 0.0075 moles / 0.05 L =0.15 M

We need 7.5 mL of the 1M stock of NaCl

6 0

3 years ago

What would be the correct answer please help!

bogdanovich [222]

Answer:

first number is the correct answer

6 0

3 years ago

Read 2 more answers

Which of the following displays the correct change in enthalpy and best describes the reaction below? 2CsCl(aq) + Na2SO4(aq) 2Na

Bond [772]

We subtract the enthalpies of the reactants from that of the products:
$2\Delta
 H(NaCl)+\Delta H(Cs_2 SO_4)-2\Delta H(CsCl)-\Delta H(Na_2 SO_4) \\ 
=2(-411)+(-1400) -2(-415)-(-1380) \\ = -12 kJ$
Since this is < 0, this is an exothermic reaction.

3 0

3 years ago

I need help solving this!

zmey [24]

Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, $CH_{4}$ .

Explanation:

Given: Mass of methane = 146.6 g

As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{146.6 g}{16.04 g/mol}\\= 9.14 mol$

The given reaction equation is as follows.

$C + 2H_{2} \rightarrow CH_{4}$

This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.

$Moles of H_{2} = \frac{9.14}{2}\\= 4.57 mol$

Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, $CH_{4}$ .

5 0

3 years ago

What is the boiling point, melting point, and triple point of bromine?

Lapatulllka [165]

Answer:

A. Boiling point = 59 °C, Melting point = -7.2°C, triple point = -7.3°C

Explanation:

3 0

3 years ago