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3 years ago

I NEED HELP ASAP WILL MARK BRAINLIEST!!!

Chemistry

1 answer:

andreev551 [17]3 years ago

7 0

The answer is b.) Levi crosses a horse with a donkey to a create a mule. The mule is stronger than both of its parents.

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How many grams of solute are needed to make 37.5 mL of 0.750 M KI solution? Round to three significant digits.

Marysya12 [62]

4.648 gm of solute is needed to make 37.5 mL of 0.750 M KI solution.

Solution:

We will start with the Molarity

$\text { Molarity }=\text { Mole of solute } \div \text { liter of solution }$

Also we know 1000 ml = 1 L

Therefore 37.5 ml by 1000ml we obtained 0.0375L

Equation for solving mole of solute

$\text { Mole of solute }=\text { Molarity } \times \text { Liters of solution }$

Now, multiply 0.750M by 0.0375

Substitute the known values in the above equation we get

$0.750 \times 0.0375=0.0281$

Also we know that Molar mass of KI is 166 g/mol

So divide the molar mass value to get the no of grams.

$0.028 \times 166=4.648$

So 4.648 gm of Solute is required for make 37.5 mL of 0.750 M KI solution.

8 0

3 years ago

What is the molarity when water is added to 4 moles of sodium chloride to make 0.5 liter of solution?

loris [4]

If 0.5 L of solution contains 4 mol
then let 1 L of solution contain x mol

⇒ (0.5 L) x = (4 mol) (1 L)

x = (4 mol · L) ÷ (0.5 L)

x = 8 mol

Thus the molarity of the Sodium Chloride solution is 8 mol / L OR 8 mol/dm³.

7 0

3 years ago

Explain why either liquids nor gases have permanent shapes

Kruka [31]

Because they can't get trapped in.

5 0

3 years ago

in a simulation mercury removal from industrial wastewater, 0.020 L of 0.10 M sodium sulfide reacts with 0.050 L of 0.010 M merc

Margarita [4]

Answer: 0.1161 grams of mercury(II) sulfide) form.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$ .....(1)

a) Molarity of $Na_2S$ solution = 0.10 M

Volume of solution = 0.020 L

Putting values in equation 1, we get:

$0.10M=\frac{\text{Moles of }Na_2S}{0.020L}\\\\\text{Moles of Na_2S}={0.10mol/L\times 0.020}=0.002mol$

$\text {Moles of}Na_2S=0.10M\times 0.020L=0.002mol$

b) Molarity of $Hg(NO_3)_2$ solution = 0.010 M

Volume of solution = 0.050 L

Putting values in equation 1, we get:

$0.010M=\frac{\text{Moles of }Hg(NO_3)_2}{0.050L}\\\\\text{Moles of }Hg(NO_3)_2={0.010mol/L\times 0.050}=0.0005mol$

$Na_2S+Hg(NO_3)_2\rightarrow HgS+2NaNO_3$

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ reacts with 1 mole of $Na_2S$

Thus 0.0005 moles of $HgNO_3$ reacts with= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg(NO_3)_2$

Thus $Hg(NO_3)_2$ is the limiting reagent and $Na_2S$ is the excess reagent.

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ forms= 1 mole of $Hg_2S$

Thus 0.0005 moles of $Hg(NO_3)_2$ forms= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg_2S$

mass of $H_2S=moles\times {\text {Molar mass}}=0.0005mol\times 232.2g/mol=0.1161g$

Thus 0.1161 grams of mercury(II) sulfide) form.

5 0

3 years ago

Show the calculation of the energy involved in condensation of 150 grams of steam at 100oC if the Heat of Vaporization for water

vaieri [72.5K]

Answer : The energy involved in condensation is, 339 kJ

Explanation :

Formula used :

$q=m\times L$

where,

q = heat required = ?

L = latent heat of vaporization of water = $2.26kJ/g$

m = mass of water = 150 g

Now put all the given values in the above formula, we get:

$q=(150g)\times (2.26kJ/g)$

$q=339kJ$

Therefore, the energy involved in condensation is, 339 kJ

4 0

3 years ago