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3 years ago

Balance the following H2 SO4 + NaOH ==== NaSO + H2O

Chemistry

1 answer:

katrin [286]3 years ago

4 0

Answer:

H2SO4 + 2NaOH ---> Na2SO4 + 2H2O

Explanation:

H2SO4 + NaOH ---> Na2SO4 + H2O

Na1 ---> 2Na

H2SO4 + 2NaOH ---> Na2SO4 + H2O

2H + 2H = 4H ----> 2H

H2SO4 + 2NaOH ---> Na2SO4 + 2H2O

4O + 2O -----> 4O + 2O

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The nucleus contains both protons and electrons True or false

Varvara68 [4.7K]

Answer: False

Explanation: The nucleus of an atom only contains the protons and neutrons.

The electrons are not found in the nucleus,

they are orbiting the nucleus in different shells.

3 0

3 years ago

I need help with balancing equations I'm doing homework and idk what to do here can you give me the answers please lol

amm1812

It's pretty easy to balance equations! Basically you want to make sure that the number of each compound is equal on both sides of the arrow.

For example number one is

Fe + H2SO4 -> Fe2(SO4)3 + H2

A 3 in front of H2SO4 because there's a subscript of 3 on the right side.

Then a 3 in front of H2 because of the previous step.

Then add a 2 in front of Fe because of the 2 subscript in Fe2(SO4)3

Then add a 1 in front of Fe2(SO4)3 because you already have an equal number of each element.

2Fe + 3H2SO4 -> 1Fe2(SO4)3 + 3H2

I hope this explanation helps! You should really do your homework because practice is everything when it comes to chemistry. You'll need to know how to do it for exams.

3 0

3 years ago

Which one of the following is not a valid expression for the rate of the reaction below? 4NH3 + 7O2 → 4NO2 + 6H2O Which one of t

Veseljchak [2.6K]

Answer : All of the above are valid expressions of the reaction rate.

Explanation :

The given rate of reaction is,

$4NH_3+7O_2\rightarrow 4NO_2+6H_2O$

The expression for rate of reaction for the reactant :

$\text{Rate of disappearance of }NH_3=-\frac{1}{4}\times \frac{d[NH_3]}{dt}$

$\text{Rate of disappearance of }O_2=-\frac{1}{7}\times \frac{d[O_2]}{dt}$

The expression for rate of reaction for the product :

$\text{Rate of formation of }NO_2=+\frac{1}{4}\times \frac{d[NO_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{6}\times \frac{d[H_2O]}{dt}$

From this we conclude that, all the options are correct.

3 0

3 years ago

Given the following data:

bagirrra123 [75]

$176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its $\Delta H$ can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with $\Delta H$ given be denoted as (1), (2), (3), and the last equation (4). Let $a$ , $b$ , and $c$ be letters such that $a \times (1) + b \times (2) + c \times (3) = (4)$ . This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, $3 + (-1) = 2$ shall resemble the number of $\text{H}_2$ left on the product side when the second equation is directly added to the third. Similarly

$\text{NH}_4 \text{Cl} \; (s)$ : $-2 \; a = 1$
$\text{NH}_3\; (g)$ : $-2 \; b = -1$
$\text{HCl} \; (g)$ : $2 \; c = -1$

Thus

$a = -1/2\\b = 1/2\\c = -1/2$ and

$-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)$

Verify this conclusion against a fourth species involved- $\text{N}_2 \; (g)$ for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

$a + b = -1/2 + 1/2 = 0$

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

$\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

3 0

3 years ago

1. write the complete chemical symbol for the ion with 14 protons 15 neutrons and 18 electrons

motikmotik

Answer:

1. $Si^{-4}$

2. $Co^{+2}$

Explanation:

An ion is formed when an atom that is said to be neutral gains or losses electrons.

It is thought that a negative ion (anion) is produced as it gains electrons and a positive ion (cation) is formed when it loses an electron.

Atomic number is the total number of protons and electrons in a neutral atom.

From the information

Protons = 14

electron = 18

Net Charge = no of proton - no of electron

= 14 - 18 = -4

Mass number = 14 + 15 = 29

Thus, the chemical symbol = $Si^{-4}$

For ion with 27 proton, 32 neutrons and 25 electrons

Net charge = 27 - 25 = +2

Mass number = 27 + 32 = 59

Thus, the chemical symbol = $Co^{+2}$

6 0

3 years ago

Balance the following H2 SO4 + NaOH ==== NaSO + H2O​

Balance the following H2 SO4 + NaOH ==== NaSO + H2O