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4 years ago

When an aqueous solution of KBr is electrolyzed,

1 answer:

3 0

KBr + H2O = KOH + Br2 + H2

So we know that we need a 2 on the left of kbr so it would equal the right Br.

=2KBr
now put a 2 in front of KOH since we need a 2k on right side

Now we put a 2 infront of H2O which wil make 4 h's and 2 o's

2h2o

now it is balanced

2KBr + 2H2O = 2KOH + Br2 + H2

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A cart is pulled by a force of 250 N at an angle of 35° above the horizontal. The cart accelerates at 1.4 m/s2. The free-body di

RideAnS [48]

The answer is 146 kg.

6 0

3 years ago

Read 2 more answers

HSiCl3+H2O=H10Si10O15+HCl

tiny-mole [99]

10 HSiCl3+ 15 H2O→H10Si10O15+ 30 HCl

Explanation:

Step 1: To make Si equal on both sides. put 10 in front of HSiCl3

10 HSiCl3+ H2O →H10Si10O15+ HCl

Step 2: By putting 30 in front of HCl, Cl can be balanced

10 HSiCl3+ H2O →H10Si10O15+ 30 HCl

Step 3: Now, balance O by putting 15 in front of H2O

10 HSiCl3+ 15 H2O→H10Si10O15+ 30 HCl

Hence the balanced equation is:

10 HSiCl3+ 15 H2O→H10Si10O15+ 30 HCl

5 0

4 years ago

PLEASE HELP ASAP!!!!!!

Alinara [238K]

C. Gravity acting on the edges of plates and convection in the mantle. So you were correct!

7 0

4 years ago

Read 2 more answers

What is the h+ concentration for an aqueous solution with poh = 3.53 at 25 ∘c?

MrMuchimi

POH+pH=14
pH=14-pOH=14-3.53=10.47
pH=-log[H⁺]
[H⁺]=10^(-pH)
[H⁺]=10^(-10.47)=3.4*10(^-11)

7 0

4 years ago

A compound that contains only carbon, hydrogen, and oxygen is 58.8% C and 9.87% H by mass. What is the empirical formula of this

Dmitry [639]

Answer: The empirical formula of the compound becomes $C_5H_{10}O_2$

Explanation:

The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.

Let the mass of the compound be 100 g

Given values:

% of C = 58.8%

% of H = 9.87%

% of O = [100 - 58.8 - 9.87] = 31.33%

Mass of C = 58.8 g

Mass of H = 9.87 g

Mass of O = 31.33 g

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Molar mass of C = 12 g/mol

Molar mass of H = 1 g/mol

Molar mass of O = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of C}=\frac{58.8g}{12g/mol}=4.9 mol$

$\text{Moles of H}=\frac{9.87g}{1g/mol}=9.87 mol$

$\text{Moles of O}=\frac{31.33g}{16g/mol}=1.96mol$

Step 2: Calculating the mole ratio of the given elements.

Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 1.96 moles

$\text{Mole fraction of C}=\frac{4.9}{1.96}=2.5$

$\text{Mole fraction of H}=\frac{9.87}{1.96}=5.03\approx 5$

$\text{Mole fraction of O}=\frac{1.96}{1.96}=1$

Converting the mole fraction into whole numbers by multiplying them with 2.

$\text{Mole fraction of C}=2.5\times 2=5$

$\text{Mole fraction of H}=5\times 2=10$

$\text{Mole fraction of O}=1\times 2=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 5 : 10 : 2

Hence, the empirical formula of the compound becomes $C_5H_{10}O_2$

8 0

3 years ago