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vlada-n [284]
2 years ago
7

How many mL of a 6 M NaOH stock solution is needed in order to prepare 500 mL of a 0.2 M NaOH solution?

Chemistry
1 answer:
WITCHER [35]2 years ago
3 0

Answer:

The right answer is "16.67 mL".

Explanation:

Given:

Molarity of NaOH,

M_1=6 \ M

M_2=0.2 \ M

Volume of NaOH,

V_1=V \ mL

V_2=500 \ mL

As we know, the equation,

⇒ M_1V_1=M_2V_2

On putting the values, we get

⇒ 6\times V=0.2\times 500

⇒ 6\times V=100

⇒        V=\frac{100}{6}

⇒            =16.67 \ mL      

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A fish company delivers 22 kg of salmon, 5.5 kg of crab and 3.48 kg of oysters to your seafood restaurant. What is the mass, in
Rainbow [258]

Answer:

Mass of sea food = 30.98 Kg

Mass of sea food in pound = 68.31 lbs

Explanation:

Salmon, crab and oysters all are sea food.

Mass of sea food = Mass of salmon + Mass of crab + mass of oyster

Mass of salmon = 22 kg

Mass of crab = 5.5 kg

Mass of oysters = 3.48 kg

Mass of sea food = Mass of salmon + Mass of crab + mass of oyster

                             = 22 + 5.5 + 3.48

                             = 30.98 Kg

1 Kg = 2.205 lbs

Therefore, 30.98 kg = 30.98 × 2.205

                                 = 68.31 lbs

8 0
2 years ago
1.72 moles of NOCI were placed in a 2.50 L reaction chamber
elena-14-01-66 [18.8K]

The equilibrium constant, Kc=0.026

<h3>Further explanation</h3>

Given

1.72 moles of NOCI

1.16 moles of NOCI  remained

2.50 L reaction chamber

Reaction

2NOCI(g) = 2NO(g) + Cl2(g).

Required

the equilibrium constant, Kc

Solution

ICE method

   2NOCI(g) = 2NO(g) + Cl2(g).

I    1.72

C   0.56           0.56         0.28

E   1.16              0.56         0.28

Molarity at equilibrium :

NOCl :

\tt \dfrac{1.16}{2.5}=0.464

NO :

\tt \dfrac{0.56}{2.5}=0.224

Cl2 :

\tt \dfrac{0.28}{2.5}=0.112

\tt Kc=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}\\\\Kc=\dfrac{0.224^2\times 0.112}{0.464^2}=0.026

4 0
3 years ago
A boat covers 2,000 meters in 30 minutes. How fast are they traveling
brilliants [131]
1+1 is 2 hope this helps
3 0
3 years ago
Sodium hydroxide (NaOH) and hydrochloric acid (HCl) combine to make table salt (NaCl) and water (H2O). Which equation is balance
Art [367]

The equation that is balanced among the given options is the first equation NaOH + HCl → NaCl + H2O

From the question,

We are to determine which of the options gives a balanced equation for the reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl) to make table salt (NaCl) and water (H₂O)

To do this, we will check which of the given equations is balanced, by checking the equations one after the other.

First, we will define what is meant by balanced equation.

Balanced equations are those whose coefficients result in equal numbers of atoms for each element in the reactants and products

  • For NaOH + HCl → NaCl + H2O

On the reactants side

Na = 1

O = 1

H = 2

Cl = 1

On the products side

Na = 1

O = 1

H = 2

Cl = 1

Since equal number of atoms of each element are present at both the reactants and products sides, then the equation is balanced

  • For 2NaOH + 2HCl → 2NaCl + H2O

On the reactants side

Na = 2

O = 2

H = 4

Cl = 1

On the products side

Na = 2

O = 1

H = 2

Cl = 2

The number of atoms of each element present at the reactants side and that present at products side are not equal.

∴ The equation is not balanced

  • For 2NaOH + HCl → NaCl + H2O

On the reactants side

Na = 2

O = 2

H = 3

Cl = 1

On the products side

Na = 1

O = 1

H = 2

Cl = 1

The number of atoms of each element present at the reactants side and that present at products side are not equal.

∴ The equation is not balanced

  • For NaOH + 2HCl → NaCl + H2O

On the reactants side

Na = 1

O = 1

H = 2

Cl = 2

On the products side

Na = 1

O = 1

H = 2

Cl = 1

The number of atoms of each element present at the reactants side and that present at products side are not equal.

∴ The equation is not balanced

Hence, the equation that is balanced is the first equation NaOH + HCl → NaCl + H2O

Learn more here: brainly.com/question/17015942

6 0
1 year ago
Read 2 more answers
Calculate the amount (in grams) of kcl present in 75.0 ml of 2.10 m kcl
Lera25 [3.4K]
V = 75 mL = 0,075 L = 0,075 dm³
C = 2.1M
n = ?
---------------
C = n/V
n = C×V
n = 2.1×0,075
n = 0,1575 mol
--------
mKCl: 39+35.5 = 74,5 g/mol

74,5g --------- 1 mol
Xg ------------- 0,1575 mol
X = 74,5×0,1575
X = 11,73375g KCl

:•)
5 0
3 years ago
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