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Anon25 [30]
3 years ago
5

What will litmus paper turn when it reacts with sodium nitrate?​

Chemistry
1 answer:
castortr0y [4]3 years ago
6 0
A solution of sodium nitrate will get make litmus paper wet. A solution of NaNO3 is neutral. Therefore, there is no change in color.
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An experimental spacecraft consumes a special fuel at a rate of 372 L/min. The density of the fuel is 0.730 g/mL and the standar
Karolina [17]

Explanation:

First, we will calculate fuel consumption is as follows.

         372 L/min \times 1000 ml/L \times 0.730 g/ml \times \frac{1}{60} min/s

           = 4526 g/s

Now, we will calculate the power as follows.

        Power = Fuel consumption rate × -enthalpy of combustion

                    = 4526 g/s \times -26.5 kJ/g

                    = 1.19 \times 10^{5} kW

Thus, we can conclude that maximum power (in units of kilowatts) that can be produced by this spacecraft is 1.19 \times 10^{5} kW.

5 0
3 years ago
Find the volume of a gas at standard pressure if its volume at 1.9 atm is 80 ml?
kogti [31]

Answer:

1.5 × 10² mL

Explanation:

Step 1: Given data

  • Initial pressure of the gas (P₁): 1.9 atm
  • Initial volume of the gas (V₁): 80 mL
  • Final pressure of the gas (P₂): 1.0 atm (standard pressure)
  • Final volume of the gas (V₂): ?

Step 2: Calculate the final volume of the gas

For an ideal gas, we can calculate the final volume of the gas using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁/P₂

V₂ = 1.9 atm × 80 mL/1.0 atm

V₂ = 1.5 × 10² mL

Since the pressure decreased, the volume of the gas increased.

6 0
2 years ago
How many electrons can occupy the 4p sublevel of a neutral atom of bromine
VMariaS [17]

Answer:

6

Explanation

every p sublevel holds up to 6 electrons

so the 4p sublevel  must hold up to 6 electonsr

4 0
2 years ago
A sample of gas has a density of 0.53 g/L at 225 K and under a pressure of 108.8 kPa. Find the density of the gas at 345 K under
sukhopar [10]

Answer:

\rho _2=0.22g/L

Explanation:

Hello!

In this case, since we are considering an gas, which can be considered as idea, we can write the ideal gas equation in order to write it in terms of density rather than moles and volume:

PV=nRT\\\\PV=\frac{m}{MM} RT\\\\P*MM=\frac{m}{V} RT\\\\P*MM=\rho RT

Whereas MM is the molar mass of the gas. Now, since we can identify the initial and final states, we can cancel out R and MM since they remain the same:

\frac{P_1*MM}{P_2*MM} =\frac{\rho _1RT_1}{\rho _2RT_2} \\\\\frac{P_1}{P_2} =\frac{\rho _1T_1}{\rho _2T_2}

It means we can compute the final density as shown below:

\rho _2=\frac{\rho _1T_1P_2}{P_1T_2}

Now, we plug in to obtain:

\rho _2=\frac{0.53g/L*225K*68.3kPa}{345K*108.8kPa}\\\\\rho _2=0.22g/L

Regards!

8 0
2 years ago
What should the IUPAC name for a binary covalent compound lack ?
zavuch27 [327]
The answer to the question is "B. Roman Numerals"
5 0
3 years ago
Read 2 more answers
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