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fgiga [73]
3 years ago
8

300 moles of sodium nitrite are needed for a reaction. the solution is 0.450 m. how many ml are needed?

Chemistry
1 answer:
miv72 [106K]3 years ago
4 0

Answer:

\boxed{\text{667 000 mL}}

Explanation:

\text{Molar concentration} = \dfrac{\text{moles}}{\text{litres}}\\\\c = \dfrac{n}{v}

Data:

c = 0.450 mol·L⁻¹

n = 300 mol

Calculation:

\begin{array}{rcl}0.450 & = & \dfrac{300}{V}\\\\0.450V& = & 300\\\\V & = & \dfrac{300}{0.450}\\\\V & = & \text{667 L} =\textbf{667 000 mL}\end{array}\\\text{The volume of solution needed is }\boxed{\textbf{667 000 mL}}

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A scientist obtains a set of measurements that are very close to one another. These can be said to be:
pickupchik [31]

Answer:

I believe this is a K-12 test question. If the answers below are what you have on your test . . .

- Precise

- Accurate

- Identical

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Then the answer is <u>precise</u>.

3 0
2 years ago
The theory general relativity was discovered by who
svetoff [14.1K]
<span>The theory general relativity was discovered by Albert Einstein </span>
4 0
2 years ago
How to find the final temperature
sveta [45]

Answer:

The final temperature will be "12.37°".

Explanation:

The given values are:

mass,

m = 0.125 kg

Initial temperature,

c = 22.0°C

Time,

Δt = 4.5 min

As we know,

⇒  q=mc \Delta t

On putting the estimated values, we get

⇒     =0.125\times 22.0\times 4.5

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8 0
2 years ago
Astroturf is a durable artificial surface used to cover athletic fields. A soccer field 0.06214- mile-long by 253 ft wide is cov
love history [14]

Answer:

The weight of the Astroturf is 179,684.31 Newtons.

Explanation:

Length of a soccer field = 0.06214 mile = 328.0992 feet

(1 mile = 5280 feet)

Breadth of a soccer field  = 253 feet

Length of a Astroturf which soccer field is to be covered, l = 328.0992 feet

Breadth of a Astroturf which soccer field is to be covered ,b = 253 feet

Thickness of a Astroturf with which soccer field is to be covered = h

h = ½ inch = 0.5 inch = 0.041665 feet

(1 inches = 0.08333 feet)

Volume of the Astroturf ,V= l × b × h

V=328.0992 ft\times 253 ft\times 0.041665 ft=3,458.574 ft^3

Mass of the Astroturf = m

Density of the Astroturf = d = 187 oz/ft^3

d=\frac{m}{V}

m=d\times V= 187 oz/ft^3\times 3,458.574 ft^3=646,753.35 oz

1 oz = 0.0283495 kg

m=646,743.35 oz=646,743.35\times 0.0283495 kg=18,335.13 kg

Weight of the Astroturf = W

W = mg

=W=18,335.13 kg\times 9.8 m/s^2=179,684.31 N

The weight of the Astroturf is 179,684.31 Newtons.

8 0
3 years ago
with strong heating calcium carbonate undergoes thermal decomposition how many mole of CaCO3 are there in 50g of calcium carbona
aev [14]
Ca=40
C=12
O=16
1 mole of CaCO3 has 100 grams
So 50 grams is 0.5 mole
6 0
2 years ago
Read 2 more answers
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