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svlad2 [7]
3 years ago
8

Write an equation and solve the problem.

Chemistry
1 answer:
VMariaS [17]3 years ago
4 0
Dexter already has 4 and he gets 38 so 4+38=42. Than 42/6=7. 7 plates on each table.
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A student obtains the following data: Mass of empty, dry graduated cylinder: 21.577 g Volume added of NaCl solution: 4.602 mL Ma
klasskru [66]

<u>Answer:</u> The density of NaCl solution is 3.930 g/mL

<u>Explanation:</u>

We are given:

Mass of cylinder, m_1 = 21.577 g

Mass of NaCl and cylinder combined, M = 39.664 g

Mass of NaCl, m_2 = (M-m_1)=(39.664-21.577)g=18.087g

To calculate density of a substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

We are given:

Mass of NaCl = 18.087 g

Volume of NaCl solution = 4.602 mL

Putting values in above equation, we get:

\text{Density of NaCl}=\frac{18.087g}{4.602mL}\\\\\text{Density of NaCl}=3.930g/mL

Hence, the density of NaCl solution is 3.930 g/mL

7 0
3 years ago
I need chemistry help! How would I set up these problems?
Bond [772]

Answer: -

1) 8.33 minutes

2) 118.39 in/ s

180.43 m/min

10.83 km/ hr

Explanation: -

Speed of light = 3 x 10⁸ m/s

Distance of the earth from the sun= 93 million miles

We know 1 million = 1,000,000

Also 1 mile = 1609 m

Distance of the earth from the sun= 93 million miles

= 93,000,000 miles.

= 1.5 x 10^{11} m

Time taken = \frac{Distance}{Speed}

= \frac{1.5 x [tex] 10^{11} m}{3 x 10⁸ m/s} [/tex]

= 500 s

= 500/ 60

= 8.33 minutes

2) Distance = 1 mile = 63360 inches

Time taken = 8.92 min

= 8.92 x 60

= 535.2 s

Speed = \frac{distance}{time}

= \frac{63360 inches}{535.2 s}

= 118.39 in/ s

Distance = 1 mile = 63360 inches = 63360 x 2.54 cm = 63360 x 2.54 x 10^{-2} m

Time taken = 8.92 min

Speed = \frac{distance}{time}

= \frac{63360 x 2.54 x [tex] 10^{-2} m}{8.92 min} [/tex]

= 180.43 m/ min

1 m = 10⁻³ Km

1 min = 1/60 hour

1 m /min = 10⁻³ km/ \frac{1}{60 hour}

= 60/1000

=0.06 km/hr

180.43 m / min = 180 x 0.06 km / hr

= 10.93 km / hr

4 0
3 years ago
What is the limiting factor in determining the accumulation of siliceous ooze/calcareous ooze, respectively?
zhannawk [14.2K]

Answer:

productivity and water depth

Explanation:

The productivity and the depth of water are both equally important as it directly affects the accumulation of biogenic sediments such as the siliceous ooze and calcareous ooze. In the equator and the coastal upwelling areas, and at the site of divergence of oceans, there occurs a high rate and amount of productivity, and these are considered to be the primary productivity.  

The siliceous oozes are a good indicator of extensively high productivity in comparison to the carbonate oozes. The main reason behind this is that the silica can be easily dissolved in the surface water. On the other hand, the carbonates dissolve at a relatively lower ocean water depth, so there requires a high amount of surface productivity in order to allow these siliceous oozes to reach the ocean bottom.

Thus, the water depth and productivity, both are considered as the limiting factor in determining the accumulation of biogenic oozes.

8 0
3 years ago
What is the balenced formula for C3H8+O2
balandron [24]

C3H8  +   5O2  ----->   3CO2  +   4H2O

This is propane  burning in air or pure oxygen.

4 0
3 years ago
For a colligative property such as freezing point depression, :________
neonofarm [45]
A possible answer is b
5 0
3 years ago
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