Which of the choices below is an empirical formula

Calculate the joules of heat needed to cool 159 g of water at 100 c to water at 65c

Butoxors [25]

Answer:

John is buying shirts for his softball team. He will pay a one-time processing fee of $27.50 and $12.75 per each shirt ordered. Which equation can be used to find y, the total cost to buy xshirts? John is buying shirts for his softball team. He will pay a one-time processing fee of $27.50 and $12John is buying shirts for his softball team. He will pay a one-time processing fee of $27.50 and $12.75 per each shirt ordered. Which equation can be used to find y, the total cost to buy xshirts? xdsz.75 per each shirt ordered. Which equation can be used to find y, the total cost to buy xshirts? John is buying shirts for his softball team. He will pay a one-time processing fee of $27.50 and $12.75 per each shirt ordered. Which equation can be used to find y, the total cost to buy xshirts? John is buying shirts for his softball team. He will pay a one-time processing fee of $27.50 and $12.75 per each shirt ordered. Which equation can be used to find y, the total cost to buy xshirts?

Explanation:

4 0

2 years ago

HELP......Percussion instruments are musical instruments that make noise when they are hit with an object. Which of these are pe

valkas [14]

Answer:

I believe that the percussion instruments are the drum and cymbal.

Explanation:

4 0

3 years ago

How do you solve for density

vodka [1.7K]

The Density Calculator uses the formula p=m/V, or density (p) is equal to mass (m) divided by volume (V). The calculator can use any two of the values to calculate the third. Density is defined as mass per unit volume.

5 0

3 years ago

Read 2 more answers

El agua del mar contiene aproximadamente un 3,0 % m/v de sal (NaCl, 58,44 g/mol), (asuma que es la única fuente de cloruros) si

Alchen [17]

Answer:

s = 4.41 g/L.

Explanation:

¡Hola!

En este caso, considerando el escenario dado, se hace necesario para nosotros saber que la posible reacción de disociación la experimenta el cloruro de plomo (II) como se muestra a continuación:

$PbCl_2(s)\rightleftharpoons 2Cl^-(aq)+Pb^{2+}(aq)$

Lo cual hace que la expresión de equilibrio se calcule como:

$Ksp=[Pb^{2+}][Cl^-]^2$

Y que en términos de la solubilidad molar, s, se resuelve como:

$1.6x10^{-5}=s(2s)^2\\\\1.6x10^{-5}=4s^3\\\\s=\sqrt[3]{\frac{1.6x10^{-5}}{4} } \\\\s=0.0159molPbCl_2/L$

Ahora, convertimos este valor a g/L al multiplicarlo por la masa molar del cloruro de plomo (II):

$s=0.0159molPbCl_2/L*\frac{278.1gmolPbCl_2}{1molmolPbCl_2} \\\\s=4.41g/L$

¡Saludos!

7 0

3 years ago

How many moles of sodium carbonate are contained by 57.3g of sodium carbonate

Lady_Fox [76]

Answer:

$\boxed {\boxed {\sf 0.541 \ mol \ Na_2CO_3}}$

Explanation:

We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.

Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.

We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.

Na: 22.9897693 g/mol
C: 12.011 g/mol
O: 15.999 g/mol

Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.

Na₂ = 22.9897693 * 2= 45.9795386 g/mol
O₃ = 15.999 * 3= 47.997 g/mol
Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol

We will convert using dimensional analysis. Set up a ratio using the molar mass.

$\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}$

We are converting 57.3 grams to moles, so we multiply by this value.

$57.3 \ g \ Na_2CO_3} *\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}$

Flip the ratio so the units of grams of sodium carbonate cancel.

$57.3 \ g \ Na_2CO_3} *\frac {1 \ mol \ Na_2CO_3}{105.9875386 \ g \ Na_2CO_3}$

$57.3 } *\frac {1 \ mol \ Na_2CO_3}{105.9875386 }$

$\frac {57.3 }{105.9875386 } \ mol \ Na_2CO_3$

$0.5406295944 \ mol \ Na_2CO_3$

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

$0.541 \ mol \ Na_2CO_3$

There are approximately <u>0.541 moles of sodium carbonate</u> in 57.3 grams.

6 0

3 years ago