Answer: So if you had 570 cm of ribbon, then 570%2F8.5=67.05 which means that about 67 students can do the experiment (round down to the nearest whole number).
Explanation: If you had 8.5 cm of ribbon, then only 8.5%2F8.5=1 student can do the experiment. If you had 17 cm of ribbon, then 17%2F8.5=2 students can do the experiment.
Answer:
Some bacteria like <em><u>rhizobium</u></em> and <u><em>blue green algae</em></u> are able to fix nitrogen gas from the atmosphere to enrich the soil with nitrogen compounds and increase its fertility. The nitrogen-fixing bacteria and blue green algae are called <u><em>biological nitrogen fixers.</em></u>
Answer:
1.14 × 10³ mL
Explanation:
Step 1: Given data
- Initial volume of the gas (V₁): 656.0 mL
- Initial pressure of the gas (P₁): 0.884 atm
- Final volume of the gas (V₂): ?
- Final pressure of the gas (P₂): 0.510 atm
Step 2: Calculate the final volume of the gas
If we assume ideal behavior, we can calculate the final volume of the gas using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁/P₂
V₂ = 0.884 atm × 656.0 mL/0.510 atm = 1.14 × 10³ mL
Answer:
—96.03°C
Explanation:
We'll begin by writing out the information provided by the question. This includes:
Number of mole (n) = 0.645 mole
Volume (V) = 2.00 L
Pressure (P) = 4.68 atm
Temperature (T) =?
Recall: that the gas constant = 0.082atm.L/Kmol
With the ideal gas equation PV = nRT, the temperature of the gas can be obtained as follow:
PV = nRT
4.68 x 2 = 0.645 x 0.082 x T
Divide both side 0.645 x 0.082
T = (4.68 x 2) /(0.645 x 0.082)
T = 176.97 K
Now, We can also express the temperature obtained in celsius as shown below:
Temperature (celsius) = temperature (Kelvin) - 273
Temperature (celsius) = 176.97 - 273
Temperature (celsius) = —96.03°C
The temperature of the Neon gas is
—96.03°C
Answer:
We need 4.28 grams of sodium formate
Explanation:
<u>Step 1:</u> Data given
MW of sodium formate = 68.01 g/mol
Volume of 0.42 mol/L formic acid = 150 mL = 0.150 L
pH = 3.74
Ka = 0.00018
<u>Step 2:</u> Calculate [base)
3.74 = -log(0.00018) + log [base]/[acid]
0 = log [base]/[acid]
0 = log [base] / 0.42
10^0 = 1 = [base]/0.42 M
[base] = 0.42 M
<u>Step 3:</u> Calculate moles of sodium formate:
Moles sodium formate = molarity * volume
Moles of sodium formate = 0.42 M * 0.150 L = 0.063 moles
<u>Step 4:</u> Calculate mass of sodium formate:
Mass sodium formate = moles sodium formate * Molar mass sodium formate
Mass sodium formate = 0.063 mol * 68.01 g/mol
Mass sodium formate = 4.28 grams
We need 4.28 grams of sodium formate
