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Dimas [21]
3 years ago
7

If you had a bottle that contained 5.69x1024 molecules of water, how many

Chemistry
1 answer:
Valentin [98]3 years ago
6 0
Yeah what the other person said ^
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A topographical map of a section of Charleston, SC is shown. What feature is located at the dot marked with an X? A) the center
Effectus [21]
It seems that you have missed the given image to answer this question. But anyway, I found it and got the answer. Based on the  topographical map of a section of Charleston, SC, the feature that is <span>located at the dot marked with an X is the high point of a hill. The answer would be option D.</span>
6 0
3 years ago
What is the unknown metal if the temperature of a beaker of 100ml of water was raised 17c to 19 c when 21 grams of the metal at
horrorfan [7]

Answer:

The metal has a heat capacity of 0.385 J/g°C

This metal is copper.

Explanation:

<u>Step 1</u>: Data given

Mass of the metal = 21 grams

Volume of water = 100 mL

 ⇒ mass of water = density * volume = 1g/mL * 100 mL = 100 grams

Initial temperature of metal = 122.5 °C

Initial temperature of water = 17°C

Final temperature of water and the metal = 19 °C

Heat capacity of water = 4.184 J/g°C

<u />

<u>Step 2: </u>Calculate the specific heat capacity

Heat lost by the metal = heat won by water

Qmetal = -Qwater

Q = m*c*ΔT

m(metal) * c(metal) * ΔT(metal) = - m(water) * c(water) * ΔT(water)

21 grams * c(metal) *(19-122.5) = -100 * 4.184 * (19-17)

-2173.5 *c(metal) = -836.8

c(metal) = 0.385 J/g°C

The metal has a heat capacity of 0.385 J/g°C

This metal is copper.

4 0
4 years ago
Which of the following are used to make predictions?
tatyana61 [14]

Answer:

Hypotheses

Explanation:

5 0
2 years ago
Realiza los cálculos para determinar la cantidad de KOH 90%, que se necesita para preparar 100 ml de solución 1N.
attashe74 [19]

Answer:

6.23 KOH 90% son necesarios

Explanation:

Una solución 1N de KOH requiere 1equivalente (En KOH, 1eq = 1mol) por cada litro de solución.

Para responder esta pregunta se requiere hallar los equivalentes = Moles de KOH para preparar 100mL = 0.100L de una solución 1N. Haciendo uso de la masa molar de KOH y del porcentaje de pureza del KOH se pueden calcular los gramos requeridos para preparar la solución así:

<em>Equivalentes KOH:</em>

0.100L * (1eq / L) = 0.100eq = 0.100moles

<em>Gramos KOH -Masa molar: 56.1056g/mol-:</em>

0.100moles * (56.1056g/mol) = 5.61 KOH se requieren

<em>KOH 90%:</em>

5.61g KOH * (100g KOH 90% / 90g KOH) =

<h3>6.23 KOH 90% son necesarios</h3>
8 0
3 years ago
PLEASE HELP ASAP I NEED IT
Lapatulllka [165]

the answer is true the first one

3 0
3 years ago
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