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Anastaziya [24]
3 years ago
14

What will be the product of the following acid-base reaction? H3O+(aq) + OH-(aq) ?

Chemistry
1 answer:
Rashid [163]3 years ago
8 0
H3O+ + OH- = 2 H2O

H3O+ is acid (proton donor)
OH- is base (proton receptor)
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On which of the following objects will the earth exert a greater gravitational pull than on a baseball?
dem82 [27]

Answer:

b.

Explanation: brainliest

5 0
2 years ago
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Consider the chemical equation. CuCl2 + 2NaNO3 mc023-1.jpg Cu(NO3)2 + 2NaCl What is the percent yield of NaCl if 31.0 g of CuCl2
lana [24]
CuCl2 + 2NaNO3 ---->  Cu(NO3)2 + 2NaCl

using molar masses:-
Theoretical yields:-
63.54 + 2(35.45) g  of CuCl2  produces  2(22.98 + 35.45) g of NaCl
    134.44  g .................................................... 116.86 g
       31.0 g ....................................................31.0 * 116.86 /134.44=26.95g  
 
 So percentage yield is 21.2* 100 / 26.95    =  78.7%  to nearest tenth                                                                                                              


5 0
3 years ago
Read 2 more answers
What sample size (grams) of Na3PO4 (FW 164.00) known to be 50.00% pure should be used to consume exactly 40.00 mL of 0.1000 M HC
Mkey [24]

Answer:

0.109 g.

Explanation:

Equation of the reaction:

Na3PO4 + 3HCl --> 3NaCl + H3PO4

Number of moles of HCl = molar concentration × volume

= 0.1 × 0.04

= 0.004 mol.

By stoichiometry, 1 mole of Na3PO4 neutralises 3 moles of HCl. Therefore, number of moles of Na3PO4 = 0.004/3

= 0.0013 mol

Mass of Na3PO4 = molar mass × number of moles

= 0.0013 × 164

= 0.219 g

Since 50% of Na3PO4 was present in the sample. Let 100 g be the total mass of the substance

= 0.219 × 50 g/100 g

= 0.109 g.

3 0
3 years ago
A smaple of 1cm cuvette gives an absorbance reading of 0.558. if they absorptivity for this sample is 15000l/(mol.cm), what is t
Mazyrski [523]
To solve this problem, we use Beer's Law: A= ε.l.c
A is the absorbance- 0,558
<span>ε is</span> the molar absorptivity- is <span>15000 </span><span><span>L⋅mol-1</span><span>cm-1</span></span>
<span>l is </span>the length of the cuvette- 1 cm
<span>c is</span> the molar concentration

Applying the formula,
0,558= 15000 x 1 x c
0,558/15000= c
c= <span>3.72×<span>10⁻⁵ </span> <span>mol⋅L<span>⁻¹</span></span></span>
<span />
3 0
3 years ago
1. If I have 45 L of He in a balloon at 25 degrees celsius and increase the temperature of the
Greeley [361]

Use Charles' Law: V1/T1 = V2/T2. We assume the pressure and mass of the helium is constant. The units for temperature must be in Kelvin to use this equation (x °C = x + 273.15 K).

We want to solve for the new volume after the temperature is increased from 25 °C (298.15 K) to 55 °C (328.15 K). Since the volume and temperature of a gas at a constant pressure are directly proportional to each other, we should expect the new volume of the balloon to be greater than the initial 45 L.

Rearranging Charles' Law to solve for V2, we get V2 = V1T2/T1.  

(45 L)(328.15 K)/(298.15 K) = 49.5 ≈ 50 L (if we're considering sig figs).

7 0
3 years ago
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