What is the concentration (M) of sodium ions in 4.57 L of a 0.398 M Na3PO4 solution?

The level of a liquid (i.e. distilled water) in laboratory glassware is determined by sight. The level corresponds to a specific

Sati [7]

Answer:

c. when the meniscus is between marks, estimate the reading to the nearest tenth of a division (ie, estimate to 0.01 mL if buret has 0.1 ml divisions)

Explanation:

When we put a liquid in a pipette, we can see that the surface of that liquid will have a curve. This curve is called the meniscus and it has the correct volume that an aqueous solution has. However, for this, it is necessary that the meniscus is observed at eye level and that its edges and center are clearly visible.

7 0

3 years ago

The electron configuration for chromium is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 5 4 s 1 instead of 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3

Bond [772]

B. Is your answer..............................

8 0

4 years ago

Read 2 more answers

Para formar bronce, se mezclan 150g de cobre a 1100°C y 35g de estaño a 560°C. Determine la temperatura final del sistema.

jek_recluse [69]

Answer:

La temperatura final del sistema es 1029,346 °C.

Explanation:

Asumamos que el sistema conformado por el cobre y el estaño no tiene interacciones con sus alrededores. Por la Primera Ley de la Termodinámica, el cobre cede calor al estaño con tal de alcanzar el equilibrio térmico. El cobre se encuentra inicialmente en su punto de fusión, mientras que el estaño está por encima de ese punto, de modo que la transferencia de calor es esencialmente sensible:

$m_{Cu}\cdot c_{Cu}\cdot (T-T_{Cu}) = m_{Sn}\cdot c_{Sn}\cdot (T_{Sn}-T)$

$(m_{Cu}\cdot c_{Cu} + m_{Sn}\cdot c_{Sn})\cdot T = m_{Sn}\cdot c_{Sn}\cdot T_{Sn} + m_{Cu}\cdot c_{Cu}\cdot T_{Cu}$

$T = \frac{m_{Sn}\cdot c_{Sn}\cdot T_{Sn}+m_{Cu}\cdot c_{Cu}\cdot T_{Cu}}{m_{Cu}\cdot c_{Cu}+m_{Sn}\cdot c_{Sn}}$ (1)

Donde:

$m_{Sn}$ - Masa del estaño, en gramos.

$m_{Cu}$ - Masa del cobre, en gramos.

$c_{Sn}$ - Calor específico del estaño, en calorías por gramo-grados Celsius.

$c_{Cu}$ - Calor específico del cobre, en calorías por gramo-grados Celsius.

$T_{Sn}$ - Temperatura inicial del estaño, en grados Celsius.

$T_{Cu}$ - Temperatura inicial del cobre, en grados Celsius.

Si sabemos que $m_{Cu} = 150\,g$ , $m_{Sn} = 35\,g$ , $c_{Cu} = 0,093\,\frac{cal}{g\cdot ^{\circ}C}$ , $c_{Sn} = 0,060\,\frac{cal}{g\cdot ^{\circ}C}$ , $T_{Sn} = 560\,^{\circ}C$ y $T_{Cu} = 1100\,^{\circ}C$ , entonces la temperatura final del sistema es:

$T = \frac{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (560\,^{\circ}C)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (1100\,^{\circ}C)}{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)}$

$T = 1029,346\,^{\circ}C$

La temperatura final del sistema es 1029,346 °C.

3 0

3 years ago

How many hydrogen grams can be obtained if

kipiarov [429]

hi im breanna

Answer:

The mole is simply a very large number that is used by chemists as a unit of measurement.

Explanation:

The mole is simply a very large number,

6.022

×

10

23

, that has a special property. If I have

6.022

×

10

23

hydrogen atoms, I have a mass of 1 gram of hydrogen atoms . If I have

6.022

×

10

23

H

2

molecules, I have a mass of 2 gram of hydrogen molecules. If I have

6.022

×

10

23

C

atoms, I have (approximately!) 12 grams.

The mole is thus the link between the micro world of atoms and molecules, and the macro world of grams and litres, the which we can easily measure by mass or volume. The masses for a mole of each element are given on the periodic table as the atomic weight. So, if have 12 g of

C

, I know, fairly precisely, how many atoms of carbon I have. Given this quantity, I know how many molecules of

O

2

are required to react with the

C

, which I could measure by mass or by volume.

5 0

3 years ago

you are holding a bowling ball with a mass of 6 kg at a height of 1 meter. How much gravitational potential energy does the bowl

Bumek [7]

The formula for GPE is PE=mgh, where “m” is the mass of the object, “g” is the acceleration due to gravity (~9.8 m/s^2 on Earth’s surface), and “h” is the height of the object from the ground. Therefore,

PE=mgh

PE=(6 kg)(9.8 m/s^2)(1 m)

PE=58.8 kg•m^2/s^2 or 58.8 Newtons

The GPE of the bowling ball under these conditions would be about 59 Newtons.

Hope this helps!

7 0

4 years ago