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schepotkina [342]
4 years ago
11

A crystalline solid has a high melting point and is known to be held together with covalent bonds. This solid is an example of _

____.
A. A network covalent solid
B. An ionic solid
C. A metallic solid
D. A molecular solid
Chemistry
1 answer:
OverLord2011 [107]4 years ago
5 0
Letter B is correct the answer
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Lakshmi has a sample of ammonium nitrate (NH4NO3) that has a mass of 40.10 g. She knows that the molar mass of NH4NO3 is 80.0432
avanturin [10]
The answer is 0.5010

Number of moles (n) is equal to the quotient of mass (m) and molar mass of a sample (Mr):
n = m/Mr

We have:
n = ?
m = 40.10 g
Mr = 80.0432 g/mol

n = 40.10 g : 80.0432 g/mol = 0.5010

<span>40.10 has 4 significant digits,
</span><span>80.0432 has 6 significant digits.
Since 4 is less than 6, we choose 4 </span>significant digits
5 0
3 years ago
How to draw Hess' Cycle for this question ?
NISA [10]

Answer : The standard enthalpy of formation of ethylene is, 51.8 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of C_2H_4 will be,

2C(s)+2H_2(g)\rightarrow C_2H_4(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)     \Delta H_1=-1411kJ/mole

(2) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.7kJ/mole

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.9kJ/mole

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equations, we get :

(1) 2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)     \Delta H_1=+1411kJ/mole

(2) 2C(s)+2O_2(g)\rightarrow 2CO_2(g)    \Delta H_2=2\times (-393.7kJ/mole)=-787.4kJ/mole

(3) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=2\times (-285.9kJ/mole)=-571.8kJ/mole

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(+1411kJ/mole)+(-787.4kJ/mole)+(-571.8kJ/mole)

\Delta H=51.8kJ/mole

Therefore, the standard enthalpy of formation of ethylene is, 51.8 kJ/mole

7 0
3 years ago
The study of the interaction of electrical and magnetic fields, and their interaction with matter, is called _____.
Crazy boy [7]

Answer:

The study of the interaction of electrical and magnetic fields, and their interaction with matter, is called Electromagnetism

7 0
4 years ago
How many grams of MgCO3 are required to neutralize 200. mL of stomach acid HCl, which is equivalent to 0.0465 MHCl?
tester [92]
<h3>Answer:</h3>

0.392 g

<h3>Explanation:</h3>

We are given the following;

Volume of HCl is 200 mL

Molarity of HCl is 0.0465M

We are required  to calculate the mass of MgCO₃ required

<h3>Step 1: Write the balanced equation for the reaction </h3>
  • The balanced equation for the reaction is;

MgCO₃(s) + 2HCl(aq) → MgCl₂(aq) + CO₂(g) + H₂O(l)

<h3>Step 2: Calculate the moles of HCl</h3>

When given the molarity of a compound and the volume, the number of moles can be calculated by;

Number of moles = Molarity × Volume

Therefore;

Volume of HCl = 0.0465 M × 0.2 L

                        = 0.0093 moles

<h3>Step 3: Calculating the number of moles of MgCO₃</h3>

From the equation, one mole of MgCO₃ reacts with two moles of HCl

Therefore, the mole ratio of MgCO₃ : HCl is 1 : 2

Hence, moles of MgCO₃ = Moles of HCl ÷ 2

                                         = 0.0093 moles ÷ 2

                                         = 0.00465 moles

<h3>Step 4: Mass of MgCO₃</h3>

To calculate the mass of a compound we need to multiply the molar mass of a compound with the number of moles.

Molar mass of MgCO₃ is 84.314 g/mol

Thus, Mass of MgCO₃ = 0.00465 moles × 84.314 g/mol

                                    = 0.392 g

Therefore, 0.392 g of MgCO₃ are required to neutralize the acid.

             

3 0
4 years ago
Why it is advisided to freshly prepare the ferrous sulphate solution for experiment
chubhunter [2.5K]
Here is a site my buddie has to help you. Well co-owner..
https://www.quora.com/Why-is-fresly-prepared-FeSO4-required-for-the-ring-test
5 0
3 years ago
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