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4 years ago

Which equation would represent the neutralization of sulfuric acid that was spilled on a base

1 answer:

8 0

Decide whether or not it represents an ionic compound. ... have the general formula HaXbOc, and organic acids, such as acetic acid, HC2H3O2. An acid, such ... acid, H2SO4, that can donate two protons are called diprotic, and some acids, such as phosphoric ..... can be poured on a strong base spill to neutralize the base.

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If you need to make 240 g lioh, how many grams of li3n must you react with excess water?

Savatey [412]

Answer is: 116,1 g of Li₃N.
Chemical reaction: Li₃N + 3H₂O → 3LiOH + NH₃.
m(LiOH) = 240g.
n(LiOH) = m(LiOH) ÷ M(LiOH).
n(LiOH) = 240 g ÷ 24 g/mol = 10 mol.
from chemical reaction: n(LiOH) : n(Li₃N) = 3 : 1.
10 mol : n(Li₃N) = 3 : 1.
n(Li₃N) = 3,33 mol.
m(Li₃N) = n(Li₃N) · M(Li₃N)
m(Li₃N) = 3,33 mol · 34,85 g/mol = 116,1 g.

8 0

3 years ago

Read 2 more answers

What would the coefficient of al(no3)3 be in the balanced equation?

soldi70 [24.7K]

First, you have to know what reacts with Aluminum Nitrate so that you can balance the equation and find out the stoichiometric coefficient. I found a similar question from another website as shown in the picture attached. Balancing the reaction would result to:

2Al(NO₃)₃ + 3FeCl₂ --> 3Fe(NO₃)₂ + 2AlCl₃

<em>The coefficient is 2.</em>

7 0

4 years ago

P 1. Elements X and Y react to form an ionic compound a formula XY. What could be the proton (atomic) numbers of X and Y?

larisa86 [58]

Answer:

the answer is C

Explanation:

This is because group6 elements are diatomic and when they are chemically combined their subscript 2 cancels out

6 0

2 years ago

Which of the following answers below is the correctly balanced half-reaction for the reduction of the SO4 ^-2 ion in acid soluti

Rom4ik [11]

Answer:

Explanation:

The oxidation number of Sulphur in SO4^2- is;

x + 4(-2) = -2

x - 8 = -2

x = -2 + 8

x = 6

Now,

the oxidation number of sulphur in H2SO3 is

2 (1) + x + 3(-2) = 0

2 + x -6 = 0

-4 + x = 0

x = 4

Hence, the oxidation number of sulphur changed from +6 to +4 which signifies gain of two electrons as shown in option C.

8 0

3 years ago

(a) Find the concentration of electrons and holes in a sample of germanium that has a concentration of donor atoms equal to 1015

matrenka [14]

Answer:

a) Germanium = 5.76 x 〖10〗^11 〖cm〗^(-3) , Semiconductor is n-type.

b) Silicon = 2.25 x 〖10〗^5 〖cm〗^(-3) , Semiconductor is n-type.

For clear view of the answers: Please refer to calculation 5 in the attachments section.

Explanation:

So, in order to find out the concentration of holes and electrons in a sample of germanium and silicon which have the concentration of donor atoms equals to 〖10〗^15 〖cm〗^(-3). We first need to find out the intrinsic carrier concentration of silicon and germanium at room temperature (T= 300K).

Here is the formula to calculate intrinsic carrier concentration: For calculation please refer to calculation 1:

So, till now we have calculated the intrinsic carrier concentration for germanium and silicon. Now, in this question we have been given donor concentration (N_d) (N subscript d), but if donor concentration is much greater than the intrinsic concentration then we can write: Please refer to calculation 2.

So, now we have got the concentration of electrons in both germanium and silicon. Now, we have to find out the concentration of holes in germanium and silicon (p_o). (p subscript o)

Equation to find out hole concentration: Please refer to calculation 3. and Calculation 4. in the attachment section.

Good Luck Everyone! Hope you will understand.

6 0

3 years ago