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babunello [35]
3 years ago
10

How many boxes of B would be required to make 30 grams of C?

Chemistry
1 answer:
Mashutka [201]3 years ago
3 0

Do you have a picture to go with it?

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What is the mass in grams of 8.65 mol C8H18?
aliina [53]

Answer:

m = 998 g

Explanation:

Hello there!

In this case, according to the definition of the molar mass as the mass of one mole of the compound, it is possible to state the 1 mole of C8H18 has a mass of 114.26 grams; therefore, the mass in 8.65 moles turn out to be:

m=8.65mol*\frac{114.26g}{1mol}\\\\m=998g

In agreement to the notation requirement.

Best regards!

5 0
3 years ago
a chemist adds of a calcium bromide solution to a reaction flask. calculate the mass in grams of calcium bromide the chemist has
mestny [16]

The mass of Calcium bromide added in the flask is 29.7 g.

<h3>What is Molarity? </h3>

Molarity is defined as the ratio of number of moles od solute to the number of volume of solution in litres.

Molarity = number of moles/ volume

<h3>Calculation of Moles</h3>

Number of moles = Molarity × volume

Given,

Molarity of Calcium bromide = 0.363 M

Volume of Calcium bromide = 410 mL

= 0.410L

By substituting all the value, we get

Number of moles = 0.363 × 0.410

= 0.148 mol

As we know that,

Molar mass of Calcium bromide = 199.89 g

<h3>What is Mole? </h3>

Mole is defined as the given mass of substance to the molar mass of substance.

Given mass = Moles × Molar mass

= 0.148 × 199.89

= 29.75 g

= 29.7 g (significant digit)

Thus, we calculated that the mass of Calcium bromide added in the flask is 29.7 g.

learn more about Molarity:

brainly.com/question/19517011

#SPJ4

DISCLAIMER:

The above question is incomplete. Below is the complete question

A chemist adds 410.0mL of a 0.363 M calcium bromide solution to a reaction flask. calculate the mass in grams of calcium bromide the chemist has added to the flask. round your answer to 3 significant digits.

4 0
1 year ago
Atoms A and X are fictional atoms. Suppose that the standard potential for the reduction of X^2+ is +0.51 V, and the standard po
IRINA_888 [86]

If you are given the standard potential for the reduction of X^2+ is +0.51 V, and the standard potential for the reduction of A^2+ is -0.33, just add the two. The standard potential for an electrochemical cell with the cell is 0.18V

5 0
3 years ago
Read 2 more answers
Consider the following equilibrium: H2CO3+H2O = H3O+HCO3^-1. What is the correct equilibrium expression?
mylen [45]

Equilibrium expression is Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\

<u>Explanation:</u>

Equilibrium expression is denoted by Keq.

Keq is  the equilibrium constant that is defined as the ratio of concentration of products to the concentration of reactants each raised to the power its stoichiometric coefficients.

Example -

aA + bB = cC + dD

So, Keq = conc of product/ conc of reactant

Keq = \frac{[C]^c [D]^d}{[A]^a [B]^b}

So from the equation, H₂CO₃+H₂O = H₃O+HCO₃⁻¹

Keq = \frac{[H3O^+]^1 [HCO3^-]^1}{[H2CO3]^1 [H2O]^1}

The concentration of pure solid and liquid is considered as 1. Therefore, concentration of H2O is 1.

Thus,

Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\

Therefore, Equilibrium expression is Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\

4 0
2 years ago
En un recipiente cerrado y rígido se introdujo una mezcla gaseosa a cierta temperatura y las presiones parciales de cada gas son
valkas [14]

Answer:

Qp > Kp, por lo tanto, la presión parcial de BrF₃(g) aumenta hasta alcanzar el equilibrio.

Explanation:

Paso 1: Escribir la ecuación balanceada

BrF₃ (g) ⇌ BrF(g) + F₂(g)      Kp(T) = 64,0

Paso 2: Calcular el cociente de reacción (Qp)

Qp = pBrF × pF₂ / pBrF₃

Qp = 1,50 × 2,00 / 0,0150 = 200

Paso 3: Sacar una conclusión

Dado que Qp > Kp, la reacción se desplazará hacia la izquierda para alcanzar el equilibrio, es decir, la presión parcial de BrF₃(g) aumenta hasta alcanzar el equilibrio.

7 0
3 years ago
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