What volume of 0.130 M HNO3 (nitric acid) reacts with 39.0 mL of 0.330 M Na2 CO3 (sodium carbonate) in the following reaction? 2HNO3 (ag) + Na2 CO3 (ag) → 2NANO3 (ag) + H2 0(1) Volume = mL Submit Answer Try Another Version 10 item attempts remaining

Part A - Relating Relative Numbers of Anions and Cations to Chemical Formulas If you have an aqueous solution that contains 1.5 mol of HCI, how many moles of ions are in the solution? О 10 О 15 О 20 O 25 О 30 Submit Request Answer Provide Feedback

Remaining Time: 1 hour, 01 minute, 47 seconds. * Question Completion Status: QUESTION 2 The number of water molecules in 2.3 mg of water is O 6.02 x 1023 O 77.0 x 1023 O 7.7 x 1023 O 7.7 x 1019 none of the above QUESTION 3 A ctudent makes a colution bu discolvina 254a ofNDOH int0 450 gof water Wh. Click Save and Submit to save and submit. Click Save All Answers to save all answer ere to search DELL

You are asked to prepare a 1.000 L solution of 4.5 M you commit a user error while preparing this solution. Assumed volume Volumetric error Preparation details Added water 2.0 cm above the line, which corresponds to 8.2 mL (0.0082 L) additional solution volume C6H12O6 (glucose; molar mass = 180.16 g/mol) in a lab by dissolving 811.0 g of glucose in water. Consider the following two scenarios in whic Prepared in a beaker Prepared in a volumetric flask 1.000 L You add the glucose to a volumetric flask and then add water until it dissolves. The water bottle you are using has a worn tip, and you inadvertently add too much water such that the meniscus is above the line. The diameter of the neck of the volumetric flask is 2.29 cm. 811.0 g glucose Concentration: You decide to evaluate and compare the errors you made while preparing the solutions using the different methods. Calculate the actual concentrations of the intended 4.5 M glucose solutions prepared by each method based on their…

ogle Search S How many grams are there in 1. D How to find the Number of Aton x + Question 4 of 8 Determine the mass in grams of HCI that can react with 0.750 g of Al(OH)s according to the following reaction Al(OH):(s) + 3 HCI(aq) → AICI:(aq) + 3 H:O(aq) ADD FACTOR ANSWER RESET *( ) 78.00 0.750 6.022 x 1023 3 1.05 0.351 1 133.33 0.0288 0.117 9.60 x 10 36.46 2.25 18.02 g Al(OH); mol HCI g AICI mol AICIs mol Al(OH): g HCI mol H20 g AICIS g H:O mol AICI,

Dilution Practice How many mL of HCI, 0.1 M, will you need to prepare 0.700 L of HCL, 0.025 M? (Ans. 175 mL of aqueous HCI, 0.1 M} Write the problem set up that justifies the answer. 1 DELL 女 23 $ & 3 6. 7 t i W e r y u d f g h j

In the laboratory, a student dilutes 20.1 mL of a 6.58 M nitric acid solution to a total volume of 250.0 mL. What is the concentration of the diluted solution? Concentration = M Submit Answer Try Another Version 3 item attempts remaining

Question 7 of 10 Submit What volume in L of a 0.724M Nal solution contains 0.405 | mol of Nal ? X STARTING AMOUNT ADD FACTOR ANSWER RESET *( ) %3D 0.405 0.293 g Nal L 126.90 g Nal/mol 0.559 mol Nal 559 mL 22.99 M Nal 148.89 6.022 x 1023 1.79 0.724 Tap here or pull up for additional resources

(Incorrect) How many grams of AgCl are formed when 7.80 mL of 0.500 M AgNO3 is added to 6.25 mL of 0.300 M NH4CI? AgNO3(aq) + NH4Cl(aq) → AgCl(s) + NH4NO3(aq) 0.0304 g 0.822 (Your answer) 0.553 g 0.269 g (Correct answer) 1.61 g

Mixture 1 Reference 2 Extra Fe(NO3)3 3 Extra KSCN 4 Add Na₂SO3 5 Add NaCl 6 Add AgNO3 Volume (mL) | Volume (mL) of 0.002 M Fe(NO3)3 2.00 2.00 2.00 2.00 2.00 2.00 of 0.002 M Other Addition KSCN 2.00 2.00 2.00 2.00 2.00 2.00 N/A 0.2 mL of 0.1 M Fe(NO3)3 0.2 mL of 0.1 M KSCN Na₂SO3 NaCl 1 mL of 0.1 M AgNO3