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2 years ago

Help! What is the the correct number for the product of this particular reaction?

Chemistry

1 answer:

bija089 [108]2 years ago

6 0

Answer: synthesis, decomposition, single-displacement, double-displacement, combustion and acid-base reactions.

Explanation:

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An igneous rock that contains mostly proxene and olivene has ___.

tangare [24]

It has an ultramafic composition
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3 years ago

HS- is amphoteric; it can behave as either an acid or a base.

Tamiku [17]

Answer:

HS+Na=>NaS+1/2H2(here HS- acts as an acid)

HS-. + HCl=> H2S(g)+ Cl-(here HS- acts as a base)

7 0

3 years ago

A student takes a measured volume of 3.00 M HCl to prepare a 50.0 mL sample of 1.80 M HCI. What volume of 3.00 M HCI

frozen [14]

Answer:

30 mL VOLUME OF 3.0 M HCl SHOULD BE USED BY THE STUDENT TO MAKE A 1.80 M IN 50 mL OF HCl.

Explanation:

M1 = 3.00 M

M2 = 1.80 M

V2 = 50 .0 mL = 50 /1000 L = 0.05 L

V1 = unknown

In solving this question, we know that number of moles of a solution is equal to the molar concentration multiplied by the volume. To compare two samples, we equate both number of moles and substitute for the required component.

So we use the equation:

M1 V1 = M2 V2

V1 = M2 V2 / M1

V2 = 1.80 * 0.05 / 3.0

V2 = 0.09 /3.0

V2 = 0.03 L or 30 mL

To prepare the sample of 1.80 M HCl in 50.0 mL from a 3.0 M HCl, 30 mL volume should be used.

5 0

3 years ago

What are the properties of gas

Oksana_A [137]

Answer:

1) easy compressed

2) fills its container

3) far more space

Explanation:

5 0

3 years ago

Read 2 more answers

My professor gave me two questions to solve using the Van Der Waals Equation. She told us to solve for P and the second one we h

Fed [463]

Answer:

P=atm

$b=\frac{L}{mol}$

Explanation:

The problem give you the Van Der Waals equation:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

First we are going to solve for P:

$(P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}$

$P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}$

Then you should know all the units of each term of the equation, that is:

$P=atm$

$n=mol$

$R=\frac{L.atm}{mol.K}$

$a=atm\frac{L^{2}}{mol^{2}}$

$b=\frac{L}{mol}$

$T=K$

$V=L$

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

$P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}$

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

$P=\frac{L.atm}{L-L}-atm$

Then operate the fraction subtraction:

P= $P=\frac{L.atm-L.atm}{L}$

$P=\frac{L.atm}{L}$

And finally you can find the answer:

P=atm

Now solving for b:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

$(V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}$

Replacing units:

$b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}$

Multiplying and dividing units,(please see the second photo below), we have:

$b=\frac{L-\frac{L.atm}{atm}}{mol}$

$b=\frac{L-L}{mol}$

$b=\frac{L}{mol}$

7 0

3 years ago