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Anton [14]
3 years ago
14

I have 0.445 L of a 1.8 M NaCl solution. If I boil away some water until the volume of the solution reaches 0.250 L, what will t

he new molarity of the solution become?
Chemistry
1 answer:
solmaris [256]3 years ago
3 0

Answer:

3.2 M

Explanation:

M1V1=M2V2

1.8*.445=.801

.801/.25=3.204

Molarity = M2 = 3.2 M

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You get Ca(II)H^2

Explanation:

This is a reaction type equation calcium reacts with HA and forms a bond.

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How much heat is required to raise the temperature of 18 grams of liquid water by 25 C?
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Since the molecular weight of water is about 18 g/mol, you have about 1 mol of water. So, now we know that it takes approximately 46.2 kJ of heat energy to warm 18.02g of water from 25C to 100C and then to vaporize it into steam. Hopefully this helped!

Explanation:

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Assume that a hydrogen atom's electron has been excited to the n ???? 5 level. how many different wavelengths of light can be em
hichkok12 [17]

n = 5 --> n = 4

n = 4 --> n = 3

n = 3 --> n = 2

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<span>So we can see that there are 4 different wavelength of light that can be emitted</span>

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4 years ago
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How is iron extracted? Describe in brief​
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A box contains 104 gas molecules, 2500 of nitrogen and 7500 of argon in thermal equilibrium. The molecular weight of N2 is 28g/m
LekaFEV [45]

Answer : The temperature of this gas will be, 206.9 K

Explanation :

The expression for the kinetic energy per molecule of monoatomic gas (argon) is:

K.E_{argon}=n_{argon}\times \frac{3}{2}\times K_BT      ...........(1)

The expression for the kinetic energy per molecule of diatomic gas (nitrogen gas) is:

K.E_{nitrogen}=n_{nitrogen}\times \frac{5}{2}\times K_BT      .............(2)

The total kinetic energy of the molecule will be,

K.E_{Total}=K.E_{argon}+K.E_{nitrogen}

Now put all the expression in this, we get:

K.E_{Total}=(7500)\times \frac{3}{2}\times K_BT+(2500)\times \frac{5}{2}\times K_BT

T=\frac{K.E}{17500\times K_B}

Now put all the given values in this expression, we get:

T=\frac{5\times 10^{-17}J}{17500\times (1.381\times 10^{-23}J/K)}

T=206.9K

Therefore, the temperature of this gas will be, 206.9 K

8 0
4 years ago
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