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alexira [117]
3 years ago
10

How many significant figures are in the measurement 10,010 grams?

Chemistry
1 answer:
leva [86]3 years ago
7 0
4 sigfigs because the 0 at the end is not significant unless there is a .(period)
the 0s inbetween numbers are always significant.
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What is the value of delta Hrxn for this equation:
Ratling [72]

Answer:

The ΔHrxn for the above equation = 179 kJ/mol

Explanation:

The reaction bond enthalpies are for the reactant;

3 × N-H = 3 × 390 = 1,170 kJ/mol

2 × O=O = 2 × 502 = 1004 kJ/mol

The reaction bond enthalpies are for the product;

3 × N-O = 3 × 201 = 603 kJ/mol

3 × O-H = 3 × 464 = 1,392 kJ/mol

The ΔHrxn for the above equation is therefore;

ΔHrxn = 1,170 + 1,004 - (603 + 1,392) = 179 kJ/mol

6 0
3 years ago
If the initial temperature of a movable cylinder was 50 degrees Celsius
slega [8]

Answer:

8.45 L

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 50 °C

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 0 °C

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 50 °C

Initial temperature (T₁) = 50 °C + 273

Initial temperature (T₁) = 323 K

Final temperature (T₂) = 0 °C

Final temperature (T₂) = 0 °C + 273

Final temperature (T₂) = 273 K

Finally, we shall determine the new volume. This can be obtained as follow:

Initial temperature (T₁) = 323 K

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 273 k

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

P₁V₁ / T₁ = P₂V₂ / T₂

2 × 5 / 323 = 1 × V₂ / 273

10 / 323 = V₂ / 273

Cross multiply

323 × V₂ = 10 × 273

323 × V₂ = 2730

Divide both side by 323

V₂ = 2730 / 323

V₂ = 8.45 L

Thus, the new volume is 8.45 L

5 0
3 years ago
What mass of natural gas (ch4) must you burn to emit 272 kj of heat?
balandron [24]
CH4 + 2 O2 ---> CO2 + 2 H2O Q = 891,6 kJ / mol CH4

1 mol CH4 = 16 g

16 g ---- 891,6 kJ
x g ----- 272 kJ

x = 272 kJ × 16 g / 891,6 kJ = 4,88 g

You must burn 4,88 g of CH4.

:-) ;-)
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3 years ago
What makes it possible
docker41 [41]

Answer:

I think long roots

Explanation:

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3 years ago
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Label the parts of the respiratory system.
Evgesh-ka [11]

Answer:

1. Nose

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3. Larynx

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