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2 years ago

I NEED HELP!!OMG OMG OMG

Chemistry

2 answers:

Virty [35]2 years ago

7 0

Answer:

AHHHHHHHHHHHHHHHHHH

Explanation:

I WISH I COULD

Eva8 [605]2 years ago

7 0

Winter is purple, spring is blue, summer is orange, fall is green.

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How many moles of O₂ are needed to react completely with 35.0 mol of FeCl₃? *

galben [10]

Answer:

52.5 mol O2

Explanation:

4 FeCl3 + 6 O2 -> 2 Fe2O3 +6 Cl2

4 mol FeCl3 -> 6 mol O2

35.0 mol FeCl3 -> x

x= (35.0 mol FeCl3 * 6 mol O2)/4 mol FeCl3

x=52.5 mol O2

5 0

3 years ago

Read 2 more answers

Should people use crispr to change human traits

noname [10]

Answer:

What do YOU think?

Explanation:

This question is asking for an opinion. The word should is included.

5 0

2 years ago

If the reaction N2 (g) + 3 H2 (g) --> 2 NH3 (g) has the concentrations 1.1 M for nitrogen, 0.75 M for hydrogen and 0.25 M fo

Luba_88 [7]

Answer: The value of $K_c$ is 0.136 and is reactant favored.

Explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For the chemical reaction between carbon monoxide and hydrogen follows the equation:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

The expression for the $K_{c}$ is given as:

$K_{c}=\frac{[NH_3]^2}{[N_2][H_2]^3}$

We are given:

$[NH_3]=0.25M$

$[H_2]=0.75M$

$[N_2]=1.1M$

Putting values in above equation, we get:

$K_c=\frac{(0.25)^2}{1.1\times (0.75)^3}$

$K_c=0.135$

There are 3 conditions:

When $K_{c}>1$ ; the reaction is product favored.
When $K_{c}$ ; the reaction is reactant favored.
When $K_{c}=1$ ; the reaction is in equilibrium.

For the given reaction, the value of $K_c$ is less than 1. Thus, the reaction is reactant favored.

Hence, the value of $K_c$ is 0.136 and is reactant favored.

4 0

3 years ago

How many grams of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O?

shepuryov [24]

Explanation:

Molarity is defined as number of moles per liter of solution.

Mathematically, molarity = $\frac{no. of moles}{Volume (in L) of solution}$

It is given that molarity is 0.0800 M and volume is 50.00 mL or 0.05 L.

molarity = $\frac{no. of moles}{Volume of solution in liter}$

0.0800 M = $\frac{no. of moles}{0.05 L}$

no. of moles = 1.6 mol

Therefore, molar mass of cupric sulfate pentahydrate is 249.68 g/mol. So, calculate the mass as follows.

No. of moles = $\frac{mass in grams}{molar mass}$

mass in grams = $no. of moles \times molar mass of CuSO_{4}.5H_{2}O$

= $1.6 mol \times 249.68 g/mol$

= 399.488 g

Thus, we can conclude that 399.488 g of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O.

4 0

3 years ago

Does anyone know how to answer this or what the answer is? My sister is having a hard time figuring this out with her tutor?

ira [324]

Its number four.Hope that helps feel free to ask me questions:)

Brainliest??

8 0

3 years ago