What coefficient of 02 should be added so the number of atoms of oxygen is conserved on both sides of the reaction equation

Write a procedure for "how to crack an egg" in cooking. You must have at least 3 steps. The format of your procedure is more imp

Nataly_w [17]

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Here's the procedure to Crack an egg ~

Step 1 ~

Hold the egg in your hand and tap it on a firm surface

Step 2 ~

Gently pull the shells apart

Step 3 ~

And release the yolk in the bowl

6 0

3 years ago

Compound A reacts with Compound B to form only one product, Compound C, and it's known the usual percent yield of C in this reac

frez [133]

Given :

Compound A reacts with Compound B to form only one product, Compound C.

The usual percent yield of C in this reaction is 40%.

10.0 g of A are reacted with excess Compound B, and 6.4 g of Compound C

To Find :

The theoretical yield of C.

Solution :

We know, % yield is given by :

$\%\ yield = \dfrac{actual\ yield}{theoretical\ yield }\times 100$

Putting given values , we get :

$40 = \dfrac{6.4}{theoretical\ yield }\times 100\\\\theoretical\ yield=\dfrac{6.4\times 100}{40}\\\\theoretical\ yield=16\ g$

Therefore, theoretical yield of C is 16 g.

Hence, this is the required solution.

7 0

3 years ago

What is the specific latent heat of fusion of ice if it takes 863 kJ to convert 4.6 kg of ice into water at 0 C?

allochka39001 [22]

Answer:

$1^{f} =187.7 \frac{J}{kg}$

Explanation:

SO in order to calculate the specific latent heat of fusion, you need to remember the formula:

$1^{f} =\frac{Q}{m}$

Where $1^{f}$ representes the specific latent heart of fusion.

$Q$ represents the heat energy added, usually represented in kJ

$m$ represents the mass of the object, in kg.

Now that we have our formula we just have to put our values into the formula:

$1^{f} =\frac{Q}{m}$

$1^{f} =\frac{863kJ}{4.6kg}$

$1^{f} =187.7 \frac{J}{kg}$

SO our answer would be $1^{f} =187.7 \frac{J}{kg}$

7 0

3 years ago

What would happen if you play a note in a cold day on a flute?

Maurinko [17]

I don't know if this is the answer you are looking for but it would be flat unless the player pushed the tuning slide in.

7 0

3 years ago

Radium has a half-life of 1,620 years. In how many years will a 1 kg sample of radium decay and reduce to 0.125 kg of radium?

Tasya [4]

We can calculate years by using the half-life equation. It is expressed as:

A = Ao e^-kt

where A is the amount left at t years, Ao is the initial concentration, and k is a constant.

From the half-life data, we can calculate for k.

1/2(Ao) = Ao e^-k(1620)
k = 4.28 x 10^-4

0.125 = 1 e^-4.28 x 10^-4 (t)
t = 4259 years

7 0

3 years ago

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