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4 years ago

How do you predict products?

Chemistry

1 answer:

mamaluj [8]4 years ago

4 0

Is this a Multiple-choice exercise

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Hydrogen gas can be produced from the reaction between methane and water. Write the balanced chemical equation that represents t

jeka57 [31]

Answer:

Atoms are neither created, nor distroyed, during any chemical reaction ... Chemical reactions are represented on paper by chemical equations. For example, hydrogen gas (H2) can react (burn) with oxygen gas (O2) to form water (H20). ... Write 'balanced' equation by determining coefficients that provide equal numbers of ...

Explanation:

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3 years ago

To cook in a dry heat is. 1-Baking 2-Broiling 3-Boiling 4-Sauteing

Degger [83]

Answer:

Baking

Explanation:

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3 years ago

What is a Newton? explain please

Lina20 [59]

Answer:

it is the unit of Force

Explanation:

force is equal to Newton

5 0

3 years ago

Read 2 more answers

An air-filled balloon has a volume of 225 L at 0.940 atm and 25 °C. Soon after, the pressure changes to 0.990 atm and the temper

lorasvet [3.4K]

Answer:

The right choice is V₂ = 195.7 L

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V: is the volume of the gas in L.

n: is the no. of moles of the gas in mol.

R: is the general gas constant,

T: is the temperature of the gas in K.

If n is constant, and have different values of P, V and T:

(P₁*V₁) / T₁ = (P₂ * V₂) / T₂

Knowing that:

V₁ = 225 L , P₁ = 0.940 atm

T₁ = 25 °C + 273 = 298 K

V₂ = ??? L, P₂ = 0.990 atm

T₂ = 0 °C + 273 = 273 K

applying in the above equation

(P ₁* V₁) / T₁ = (P₂ * V₂) / T₂

(0.940 atm * 225 L) / 298 K= (0.990 atm * V₂) / 273 K

V₂ = (0.940 atm * 225 L * 273 K) / (298 K * 0.990 atm)

V₂ = 195.7 L

So, the right choice is:

V₂ = 195.7 L

7 0

3 years ago

What is the percent by mass of water in MgSO4 • 7H2O? A. 51.1% B. 195% C. 56.0% D. 21.0%

lbvjy [14]

Answer : The correct option is, (A) 51.1%

Explanation :

Mass percent : It is defined as the mass of the given component present in the total mass of the compound.

Formula used :

$\text{Mass} \%H_2O=\frac{\text{Mass of }H_2O}{\text{Mass of }MgSO_4.7H_2O}\times 100$

First we have to calculate the mass of $H_2O$ and $MgSO_4.7H_2O$ .

Mass of $H_2O$ = 18 g/mole

Mass of $7H_2O$ = 7 × 18 g/mole = 126 g/mole

Mass of $MgSO_4.7H_2O$ = 246.47 g/mole

Now put all the given values in the above formula, we get the mass percent of $H_2O$ in $MgSO_4.7H_2O$ .

$\text{Mass} \%H_2O=\frac{126g/mole}{246.47g/mole}\times 100=51.1$

Therefore, the mass percent of $H_2O$ in $MgSO_4.7H_2O$ is, 51.1%

6 0

3 years ago