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2 years ago

A higher temperature of fresh concrete results in a __________ hydration of cement.

Chemistry

1 answer:

Readme [11.4K]2 years ago

4 0

Answer:

more rapid

Explanation:

A higher temperature of fresh concrete results in a more rapid hydration of cement. This causes reduction in the setting time of the cement, also known as accelerated setting of the cement.

It also reduces the workability of the concrete; as it makes the movement of aggregates harder by reducing the lubricating effect of the cement.

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Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. If 18.1 g of Nh3 is r

Scilla [17]

I first converted the given grams of the reactants into moles, and then divided the moles by the coefficients in front of each of the reactant. The result with the smallest value will be the limiting reactant, and the value of CuO was the smallest, so it's the limiting reactant.

After figuring out which reactant is the limiting one, I took their given grams and converted it into moles, the divided it by the ratio of N2 to CuO (it's in the equation) to obtain the moles of N2, and then multiply it with the molar mass of N2 to get its mass in grams.

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1 year ago

How many hydrogen bonds can CH2O make to water

VladimirAG [237]

Hydrogen bonds are not like covalent bonds. They are nowhere near as strong and you can't think of them in terms of a definite number like a valence. Polar molecules interact with each other and hydrogen bonds are an example of this where the interaction is especially strong. In your example you could represent it like this:

H2C=O---------H-OH 

But you should remember that the H2O molecule will be exchanging constantly with others in the solvation shell of the formaldehyde molecule and these in turn will be exchanging with other H2O molecules in the bulk solution. 

Formaldehyde in aqueous solution is in equilibrium with its hydrate. 

H2C=O + H2O <-----------------> H2C(OH)2

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1 year ago

2N2H4(l) + N2O4(l) → 3N2(g) + 4H2O(g) [balanced] How many moles of N2H4 is required to produce 28.3 g of N2? Assume that all rea

JulijaS [17]

Answer: 0.67 moles of $N_2H_4$

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles of nitrogen}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{28.3}{28.02}=1mole$

$2N_2H_4(l)+N_2O_4(l)\rightarrow 3N_2(g)+4H_2O(g)$

According to stoichiometry:

3 moles of $N_2$ is produced by 2 moles of $N_2H_4$

Thus 1 mole of $N_2$ is produced by= $\frac{2}{3}\times 1=0.67moles$ of $N_2H_4$

Thus 0.67 moles of $N_2H_4$ are required to produce 28.3 g of $N_2$

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2 years ago

Crime scene investigators keep a wide variety of compounds on hand to help with identifying unknown substances they find in the

Sergio039 [100]

So 100 g of this substance has 63.57 g of carbon, 6 g of hydrogen, 9.267 of nitrogen, and 21.17 of oxygen. I need to have them all in moles (n). You can find the molar mass (M) of each element in a periodic table.

n = m/M
63.57 g C -> 63.57 g C/12.01 g/mol = 5.29 moles C
6 g H -> 6 g C/1.008 g/mol = 5.95 moles H
9.267 g N -> 9.267 g N/14.01 g/mol = 0.6615 moles N
21.17 g O -> 21.17/16.00 g/mol = 1.32 moles O

So the minimum formula has this rate:

C 5.29 H 5.95 N 0.6615 O 1.32

Now you should divide all those numbers by the smallest one (1.32):

C 4 H 4.5 N 0.5 0 1

Now it looks a lot more like a molecular formula… but we still have fractions.

Let’s multiply all numbers by 2:

C8H9N1O2

Now they are all whole numbers!

So the minimum formula is C8H9NO2

The minimum formula is not always equal to the molecular formula… but in this case, I found there is a molecular formula with this same numbers, and it’s called acetaminophen.

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2 years ago

Read 2 more answers

Air in a 0.3 m3 cylinder is initially at a pressure of 10 bar and a temperature of 330K. The cylinder is to be emptied by openin

sergiy2304 [10]

Answer:

(a) Temperature = 330 K, and mass = 0.321 kg

(b) T₂ = 171.56 K, mass = 0.32223 kg

Explanation:

For a constant temperature process we have

p₁v₁ = p₂v₂

Where p₁ = initial pressure = 10 bar = 1000000 Pa

p₂ = final pressure = 1 atm = 101325 Pa

v₁ initial volume = 0.3 m³

v₂ = final volume = unknown

From the relation we have v₂ = 2.96 m³

Therefore at constant temperature 2.93 m³ - 0.3 m³ or 2.66 m³ will be expelled from the container

Temperature = 330 K, and mass =

Also from the relation p1v1 = mRT1

We have, (1000000×0.3)/(8314×330) = 109..337 mole

For air mass

Mass = 3.171 kg

After opening we have

p2v2/(RT1) = n2 = 11.07 mol or 0.321 kg

(b) This is said to be adiabatic condition hence

Here

But cp = 29 (J/mol K).

and p₁v₁ = RT₁ therefore R = 1000000*0.3/330 = 909.1 J/mol·K

And For perfect gas γ = 1.4

Hence T₂ = 171.56 K

γ =cp/cv therefore cv=cp/γ = 29/1.4 = 20.714 (J/mol K). and R =cp-cv = 8.29 J/mol·K

Therefore p1v1/(RT1) = 109.66 moles and we have

p2v2/(R×T2) = 11.11 mole left

For air that is 0.32223 kg

5 0

2 years ago