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3 years ago

True or false: Some parasites have complex life cycles where they live in different species of hosts at different stages

Chemistry

2 answers:

Fiesta28 [93]3 years ago

7 0

Answer:

True there are a few.

Explanation:

GrogVix [38]3 years ago

6 0

Answer:

true

Explanation:

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How could you determine whether one of these two suspected compounds was identical to the unknown compound without using any for

sukhopar [10]

Answer: Use Mixture Melting Point

Explanation:

A procedure called mixture melting point would be used to determine whether or not the suspected compound is identical to the unknown.

The two suspected compounds would need to be used to create a new mixture and determine the mixtures melting point. Compare this melting points with that of the unknown compound in order to determine which one of these two suspected compounds is identical to the unknown compound.

4 0

3 years ago

What is the molar mass of (NH4)3 PO3?

vladimir1956 [14]

133.0873 g/mol

(NH4)3PO3 - molar mass

7 0

3 years ago

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In a chemical reaction, exactly 2 mol of substance A react to produce exactly 3 mol of substance B. 2 A ⟶ 3 B How many molecules

Harman [31]

Answer:

There will be produced 1.71 moles of B which contain 1.03×10²⁴ molecules

Explanation:

The example reaction is:

2A → 3B

2 moles of A produce 3 moles of B

If we have the mass of A, we convert it to moles and then, we make the rule of three: 29.2 g / 25.6g/mol = 1.14 moles

Therefore 2 moles of A produce 3 moles of B

1.14 moles of A will produce (1.14 . 3) / 2 = 1.71 moles of B are produced

Now we can determine, the number of molecules

1 mol has NA molecules (6.02×10²³)

1.71 moles have (1.71 . NA) = 1.03×10²⁴ molecules

4 0

3 years ago

I NEED THIS ASAP Someone please help me

IceJOKER [234]

Answer:

B. Four moles of water were produced from this reaction.

Explanation:

I took the test and got it correct

8 0

3 years ago

Read 2 more answers

The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol

Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration $PCl_3$ and $Cl_2$ .

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}$

$\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M$

$\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}$

$\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M$

The given equilibrium reaction is,

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initially 0.70 0.70 0

At equilibrium (0.70-x) (0.70-x) x

The expression of $K_c$ will be,

$K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}$

$K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}$

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

$49=\frac{(x)}{(0.70-x)\times (0.70-x)}$

By solving the term x, we get

$x=0.59\text{ and }0.83$

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of $PCl_3$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $Cl_2$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $PCl_5$ at equilibrium = x = 0.59 M

Therefore, the concentration of $PCl_3$ at equilibrium is 0.11 M

3 0

3 years ago