All categories

3 years ago

Why did a drop of water need to be added in order to initiate the reaction?

Chemistry

2 answers:

jeka57 [31]3 years ago

5 0

D.because the water is needed to wet the sand and keep the reaction safe

Rufina [12.5K]3 years ago

3 0

I think it’s D . If that’s not it I’m sry I tried lol

You might be interested in

Calculate the specific heat of a substance when 63j of energy are transferred as heat to an 8.0 g sample to raise it temperature

Flura [38]

The formula for energy or enthalpy is:

E = m Cp (T2 – T1)

where E is energy = 63 J, m is mass = 8 g, Cp is the specific heat, T is temperature

63 J = 8 g * Cp * (340 K – 314 K)

Cp = 0.3 J / g K

6 0

3 years ago

Read 2 more answers

If 12.1 kilograms of al2o3(s), 60.4 kilograms of naoh(l), and 60.4 kilograms of hf(g) react completely, how many kilograms of cr

Fynjy0 [20]

Answer is: 7,826 kg of cryolite.
Chemical reaction: Al₂O₃ + 6NaOH + 12HF → 2Na₃AlF₆ + 9H₂O.
m(Al₂O₃) = 12,1 kg = 12100 g.
n(Al₂O₃) = m(Al₂O₃) ÷ M(Al₂O₃).
n(Al₂O₃) = 12100 g ÷ 101,96 g/mol = 111,86 mol; limiting reactant.
m(NaOH) = 60,4 kg = 60400 g.
n(NaOH) = 60400 g ÷ 40 g/mol.
n(NaOH) = 1510 mol.
m(HF) = 60,4 kg = 60400 g.
n(HF) = 60400 g ÷ 20 g/mol = 3020 mol.
From chemical reaction: n(Al₂O₃) : n(Na₃AlF₆) = 6 : 2.
n(Na₃AlF₆) = 2 ·111,86 mol ÷ 6 = 37,28 mol.
m(Na₃AlF₆) = 37,28 mol · 209,94 g/mol.
m(Na₃AlF₆) = 7826,56 g = 7,826 kg.

7 0

3 years ago

Hc1 is the formula for .

PIT_PIT [208]

Answer:

HCl is the formula for Hydrochloric acid

Explanation:

Chemical formula is a formula of a compound showing the symbols of elements present in the compound.
Chemical formula also shows the number of atoms of each element present in a compound.
HCl is the chemical formula of hydrochloric acid. From this formula we can tell that hydrochloric acid is made up of hydrogen and chlorine elements.
The formula also shows that HCl contains 1 hydrogen atom and 1 chlorine atom.

5 0

3 years ago

Calcium carbonate is often used as an antacid. Your stomach acid is composed of HCl at a pH of 1.5. If you ate toooo much Turkey

stiks02 [169]

Answer: 0.0237 g of calcium carbonate would be required to neutralize the given amount of HCl

Explanation:

pH is defined as the negative logarithm of hydrogen ion concentration present in the solution

$pH=-\log [H^+]$ .....(1)

Given value of pH = 1.5

Putting values in equation 1:

$1.5=-\log[H^+]$

$[H^+]=10^{(-1.5)}=0.0316M$

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

$\text{Molarity of solution}=\frac{\text{Number of moles of solute}\times 1000}{\text{Volume of solution (mL)}}$ .....(2)

We are given:

Volume of solution = 15.0 mL

Molarity of HCl = 0.0316 M

Putting values in equation 2:

$0.0316=\frac{\text{Moles of HCl}\times 1000}{15.0}\\\\\text{Moles of HCl}=\frac{0.0316\times 15.0}{1000}=4.74\times 10^{-4}mol$

The chemical equation for the reaction of HCl and calcium carbonate follows:

$2HCl+CaCO_3\rightarrow H_2CO_3+CaCl_2$

By the stoichiometry of the reaction:

2 moles of HCl reacts with 1 mole of calcium carbonate

So, $4.74\times 10^{-4}mol$ of HCl will react with = $\frac{1}{2}\times 4.74\times 10^{-4}=2.37\times 10^{-4}mol$ of calcium carbonate

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of calcium carbonate = $2.37\times 10^{-4}mol$

Molar mass of calcium carbonate = 100.01 g/mol

Putting values in the above equation:

$\text{Mass of }CaCO_3=(2.37\times 10^{-4}mol)\times 100.01g/mol\\\\\text{Mass of }CaCO_3=0.0237g$

Hence, 0.0237 g of calcium carbonate would be required to neutralize the given amount of HCl

7 0

2 years ago

A 0.300 m solution of hcl is prepared by adding some 1.50 m hcl to a 500 ml volumetric flask and diluting to the mark with deion

liberstina [14]

In dilution we add distilled water to decrease the concentration of required sample from high concentration to lower concentration
The law used for dilution:
M₁V₁]Before dilution = M₂V₂] After dilution
M₁ = 1.5 M
V₁ = ?
M₂ = 0.3 M
V₂ = 500 ml
1.5 * V₁ = 0.3 * 500 ml
so V₁ = 100 ml and it completed to 500 ml using 400 ml deionized water

8 0

3 years ago