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3 years ago

Helppppppppppppppppppp????

Chemistry

2 answers:

Yuri [45]3 years ago

6 0

Answer: There are three main types of MS, relapsing primary progressive and secondary progressive

const2013 [10]3 years ago

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what is the name of first bulgarian personal computer

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How many moles of hydrogen atoms are there in 120 g of C6H12O6

lys-0071 [83]

Answer: 8.0 moles

Explanation:

0.6661 moles×12 H≈8.0 moles

5 0

3 years ago

The component that dissolves the other component is called the

zheka24 [161]

Answer:

The component that dissolves the other component is called the solvent. Solute – The component that is dissolved in the solvent is called solute

6 0

3 years ago

Read 2 more answers

172 °C = __________ K<br> (Round off to the nearest whole number)

NNADVOKAT [17]

-101, K=C+273 so in this you would subtract 273 from 172

8 0

3 years ago

In the coal-gasification process, carbon monoxide is converted to carbon dioxide via the following reaction: CO (g) H2O (g) CO2

jolli1 [7]

Answer: $K_{eq}$ at the temperature of the experiment is 0.56.

Explanation:

Moles of $CO$ = 0.35 mole

Moles of $H_2O$ = 0.40 mole

Volume of solution = 1.00 L

Initial concentration of $CO$ = $\frac{0.35mol}{1.00L}=0.35M$

Initial concentration of $H_2O$ = $\frac{0.40mol}{1.00L}=0.40M$

Equilibrium concentration of $CO$ = $\frac{0.19mol}{1.00L}=0.19M$

The given balanced equilibrium reaction is,

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

Initial conc. 0.35 M 0.40 M 0 M 0M

At eqm. conc. (0.35-x) M (0.40-x) M (x) M (x) M

Given: (0.35-x) = 0.19

x= 0.16 M

The expression for equilibrium constant for this reaction will be,

$K_{eq}=\frac{[CO_2]\times [H_2]}{[CO]\times [H_2O]}$

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

$K_{eq}=\frac{0.16\times 0.16}{(0.35-0.16)\times (0.40-0.16)}$

$K_{eq}=\frac{0.16\times 0.16}{(0.19)\times (0.24)}=0.56$

Thus $K_{eq}$ at the temperature of the experiment is 0.56.

3 0

3 years ago

If 30.0 mL of Ca(OH)2 with an unknown concentration is neutralized by 40.0 mL of 0.175 M HCl, what is the concentration of the C

Over [174]

Answer:

Concentration of Ca(OH)₂:

0.117 M.

Explanation:

How many moles of HCl is consumed?

Note the unit of concentration: moles per liter solution.

$c(\text{HCl}) = 0.175\;\text{M} = 0.175\;\text{mol}\cdot\textbf{L}^{-1}$ .

Convert milliliters to liters.

$V(\text{HCl})=40.0\;\text{mL} = 0.0400\;\text{L}$ .

$n(\text{HCl}) = c(\text{HCl})\cdot V(\text{HCl})= 0.175\;\text{mol}\cdot\text{L}^{-1} \times 0.0400\;\text{L}= 7.00\times 10^{-3}\;\text{mol}$ .

How many moles of NaOH in the solution?

Refer to the equation. The coefficient in front of Ca(OH)₂ is 1. The coefficient in front of HCl is 2. In other words, it takes two moles of HCl to neutralize one mole of Ca(OH)₂. That $7.00\times 10^{-3}\;\text{mol}$ of HCl will neutralize only half that much Ca(OH)₂.

$\displaystyle n(\text{Ca}(\text{OH})_2)=\frac{1}{2}\;n(\text{HCl}) = 3.50\times 10^{-3}\;\text{mol}$ .

What's the concentration of the Ca(OH)₂ solution?

Concentration is the number of moles of solute per unit volume.

$\displaystyle c(\text{Ca}(\text{OH})_2) = \frac{n(\text{Ca}(\text{OH})_2)}{V(\text{Ca}(\text{OH})_2)} = \frac{3.50\times 10^{-3}\;\text{mol}}{0.0300\;\text{L}}=0.117\;\text{mol}\cdot\text{L}^{-1}$ .

3 0

3 years ago