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- a. 23.25 + 625 = b. 58.175 – 45.06 = %3D c. 1000. × 2.68 =a .ull Asiacel| 10 N 141 N O A box with a volume (V=0.05 m³) lies at the bottom of a lake whose water has a density of (1*10³ kg/m³). How much force is required to lift the box, .if the mass of the box is (1000 kg) 9319.5 N O 9313.9 N O 9391.5 N O 9315.9 N O A future rectangular ship filled with oil. If the dimensions of the ship are (250 m) long, (80 m) wide, and (80 m) high. Determine how far the bottom of the ship is below the sea level? (Consider the total mass of the ship with the oil is (10.2*108 kg), and the .sea density is (1024 kg/m³))Inbc X (531 X Con X I Bala X Ans X Inbc X CHE X 101 Che X С Ч-С х b My x Unk X Sea X E I ma X app.101edu.co/# Bryant's Gmail Cascadia Canvas Lo... T GSBA Scholarship L... HOMEGROWN TRA... Learn Touch Typing... C The Science of Well... Investor360° ® Login ClickUp Reading list >> Question 7 of 40 Submit Which of the following compounds can form intermolecular hydrogen bonds? A) CH4 B) H2Se C) NH3 D) H2 E) All of these compounds can form hydrogen bonds. 10:36 AM e Type here to search 64°F 小 8/26/2021 (8)
- 1. Convert 168 nm3 to ft3. a.5.93 x 10exp-16 ft^3b.5.93 x 10exp-24 ft^3c.5.22 x10exp22 ft3d.6.00 x10exp-15 ft3O Playing a Game-Quizizz + x + + + C O a quizizz.com/join/game/U2FSDGVKX1%252FTIU7Xp3BIVNwrBmb2ip3HdTht7znvrOUUSZ%252BOm6%252BmvTKEgsg18lgVN. a GTA Student Bookmarks V Webkinz-Come in Google Slides F Google Docs A https://web.learnin. MusicFirst LMS 6 How Did the inventi Microsoft Word - C E What is the purpos 19/20 Atomic radius decreases as we move from left to right across a period. Explain why. In your explanation discuss the atom's structure: energy levels (rings), protons, and/or valence electrons. Write your response. SUBMIT Music on uS 13:42 esc C @ 2$ % & 1 4. 5 7 9 backspace %3D W e r t y i u a k C V > + 10a .ll Asiacel| 10 N 141 N O A box with a volume (V=0.05 m³) lies at the bottom of a lake whose water has a density of (1*10³ kg/m³). How much force is required to lift the box, .if the mass of the box is (1000 kg) 9319.5 N O 9313.9 N O 9391.5 N O 9315.9 N O A future rectangular ship filled with oil. If the dimensions of the ship are (250 m) long, (80 m) wide, and (80 m) high. Determine how far the bottom of the ship is below the sea level? (Consider the total mass of the ship with the oil is (10.2*108 kg), and the .sea density is (1024 kg/m³))
- a. When 90.65 and 36.969 are multiplied, the answer should have ( Enter the answer with the correct number of digits. 90.65 x 36.969 = b. When 90.65 and 36.969 are added, the answer should have v Enter the answer with the correct number of digits. 90.65+ 36.969 =Carry out each calculation and report the answer using the proper number of signifi cant fi gures. a. 53.6 × 0.41c. 65.2 ÷ 12e. 694.2 × 0.2 b. 25.825 − 3.86d. 41.0 + 9.135f. 1,045 − 1.26A 125mL flask and stopper have a mass of 79.310g. A 25.0mL sample of an unknown volatile liquid is pipetted into the flask. The flask, stopper and liquid have a total mass of 97.010g. What is the density of the unknown liquid?
- 1. A student wanted to find the density of an object of unknown composition with a volume of 5.50 0.05 cm³. The mass of the object was 19.255 0.005 g. What is the density of the object, including the absolute uncertainty? a. 3.500 ± 0.3 g cm³ b. 3.500 ± 0.03 g cm? c. 4.259 ± 0,6 g cm¹ d. 4,259 ± 0,06 g cm³The speed of light is 3 x 108 m/sec Assume: 1.6 km/mile 1.If the earth is close to Mars, we are roughly 50 million miles apart. How many minutes will it take for us to communicate with astronauts on Mars? 2.When we are at opposites in our orbits, we are close to 200 million miles apart. How many minutes will it take for us to communicate with astronauts on Mars?Find the difference between the value of x2 and the sum of x1 plus x3.; let x4 be equivalent to the sum of x1 and x3 and let x5 be the difference between x2 and x5. This calculation scheme is shown below. Make notes about any similarities or differences between the values in your notes. x4 = x1 + x3 x5 = x2 – x4 Data Analyis This section will include all data collected during the lab. Thermochemical Data Tinitial (°C) Tfinal (°C) ΔT (°C) moles NaOH qreaction (kJ) ΔHrxn Reaction 1 25.0 30.3 +5.3 0.025 -1.11 -44.4 Reaction 2 25.0 37.0 +12.0 0.025 -2.51 -100.4 Reaction 3 25.0 31.7 +6.7 0.025 -1.40 -56.1 Reaction 1: NaOH(s) → Na+(aq) + OH-(aq) + x1 kJ 1g /39.977g/mol = 0.025 moles Moles NaOH = 0.025 qsolution = (4.184 J/g °C) (50.0g) (30.3°C -25.0°C) = -1108.76 J/ 1000 qreaction (kJ) = -1.11 kJ ΔH = -1.11 kJ/ 0.025 moles ΔHrxn = -44.4 kJ/mol…