All categories

3 years ago

What is the name of the people who hid the Dead Seas Scrolls. When did they live at Qumran?

Chemistry

1 answer:

Shkiper50 [21]3 years ago

3 0

It was named the Qumran seas scrolls.

You might be interested in

What happens when a solid becomes a liquid?

Mnenie [13.5K]

All its doing is changing its physical state so it chemical state not being changed means its still the same thing it was as a solid, just now its turned into a liquid.

5 0

3 years ago

Read 2 more answers

PLEAASSEEE LOVE U GUYS ANYWAY

avanturin [10]

Answer:

ok just explanation

Explanation:

1. Cation, ion

2. octet rule

3. ion

4. anion, ion

5. cation, maybe ion?

6. Anion, maybe ion?

8 0

3 years ago

How much rust is produced with 1.5 kg of Fe reacts with water

ZanzabumX [31]

Answer:

2071g or 2.071kg of rust (Fe3O4)

Explanation:

Step 1:

The balanced equation for the reaction.

3Fe + 4H2O —> Fe3O4 + 4H2

Step 2:

Determination of the mass of Fe that reacted and the mass of the Fe3O4 produced from the balanced equation.

Molar Mass of Fe = 56g/mol

Mass of Fe from the balanced equation = 3 x 56 = 168g

Molar mass of Fe3O4 = (56x3) + (16x4) = 232g/mol

Mass of Fe3O4 from the balanced equation = 1 x 232 = 232g

Summary:

From the balanced equation above,

168g of Fe reacted and 232g of Fe3O4 was produced.

Step 3:

Determination of the mass of rust (Fe3O4) produced when 1.5kg ( i.e 1500g) of Fe reacted.

This is illustrated below:

From the balanced equation above,

168g of Fe reacted to produce 232g of Fe3O4.

Therefore, 1500g of Fe will react to produce = (1500x232)/168 = 2071g of Fe3O4.

From the calculations made above, 2071g or 2.071kg of rust (Fe3O4) is produced.

6 0

3 years ago

"how many grams of h2 are needed to produce 13.09 g of nh3

Vadim26 [7]

MH₂: 2 g/mol
mNH₃: 17 g/mol
...............................
3H₂ + N₂ ---> 2NH₃
6g......................34g

6g H₂ ---- 34g NH₃
Xg -------- 13,09g
X = (6×13,09)/34
X = 2,31g H₂

5 0

3 years ago

What is the mean free path for the molecules in an ideal gas when the pressure is 100 kPa and the temperature is 300 K given tha

sladkih [1.3K]

Answer:

The mean free path = 2.16*10^-6 m

Explanation:

Given:

Pressure of gas P = 100 kPa

Temperature T = 300 K

collision cross section, σ = 2.0*10^-20 m2

Boltzmann constant, k = 1.38*10^-23 J/K

To determine:

The mean free path, λ

Calculation:

The mean free path is related to the collision cross section by the following equation:

$\lambda =\frac{1}{n\sigma }------(1)$

where n = number density

$n = \frac{P}{kT}-----(2)$

Substituting for P, k and T in equation (2) gives:

$n = \frac{100,000 Pa}{1.38*10^{-23} J/K*300K} =2.42*10^{25}\ m^{3}$

Next, substituting for n and σ in equation (1) gives:

$\lambda =\frac{1}{2.42*10^{25}m^{-3}* .0*10^{-20}m^{2}}=2.1*10^{-6}m$

6 0

3 years ago