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Neporo4naja [7]
3 years ago
10

n the table below, write the density of each object. Then predict whether the object will float or sink in each of the fluids. W

rite “Float” or “Sink” in each empty box of the table. Object Object density Liquid density 0.5 g/mL 1.0 g/mL 2.0 g/mL 1 1.4 sink sink float 2 3 0.8 sink float float 4 5
Chemistry
1 answer:
Scrat [10]3 years ago
3 0

Answer:

0.5 g/mL----- float

1.0 g/mL---- float

2.0 g/mL-----sink

Explanation:

Those objects will float whose density is less or equal to water density due to lower mass while those object will sink whose density is greater than water because the mass of the object is greater than water. So the density of the object i. e. 0.5 g/mL and 1.0 g/mL will float in the liquid because its density is lower than water which is 1 g/mL while the density of an object i. e. 2.0 g/mL is denser than water so it will sink.

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Which formulas represent two polar molecules?
Tresset [83]
The correct answer is shown in option 3. Water and hydrochloric acid are polar molecules. These molecules are polar because of the presence of bonds that are partially ionic or polar covalent bonds. Other examples are hydrogen fluoride and ammonia. 
7 0
4 years ago
A chemical reaction was carried out three times. The mass of the product was 8.93 g for the first trial, 8.94 g for the second t
joja [24]

The three mass value measure are precise mass

<u>explanation</u>

precise mass is term use to describe data from experiment that have  been repeated several times. An experiment that yield tightly grouped  set of data it has a high precision.  8.93 , 8.94  and 8.92 are precise mass since they have repeated severally

6 0
4 years ago
The reducing agent in this catalytic hydrogenation reaction was molecular hydrogen (H2), which was produced in situ (in the reac
creativ13 [48]

Answer:

a)  After the balloon inflated after 440 uL of dropwise due to the reaction of 1-Decene and the solution in the conical vial. b) 4NaBH_{4} + 2HCl + 7H_{2}O ⇒ 16H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7} c) No H_{2} was not the limiting reactant.

Explanation:

Generally, hydrogenation is the chemical reaction between a compound or element and molecular hydrogen in the presence of catalysts such as platinum.

a) After the balloon inflated after 440 uL of dropwise 1-Decene solution was added due to the reaction between 1-Decene and the solution in the conical vial.

b)  4NaBH_{4} + 2HCl + 7H_{2}O ⇒ 16H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}

c) H_{2} was not the limiting reactant based on the mol to mol ratio of H_{2} and decane which is 1:1. Therefore, if 0.8 mol of decane was produced then 0.8 mol of H_{2} would also be produced.

4 0
3 years ago
A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci
Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}

The ICE table is then shown as:

                               C_3H_6ClCO_2H_{(aq)}  \ \ \ \ \to  \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ +  \ \ \ \ H^+_{(aq)}

Initial   (M)                     1.8                                       0                               0

Change  (M)                   - x                                     + x                           + x

Equilibrium   (M)            (1.8 -x)                                  x                              x

K_a  = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}

where ;

K_a = 3.02*10^{-5}

3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}

Since the value for K_a is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

3.02*10^{-5} *(1.8) = {(x)(x)}

5.436*10^{-5}= {(x^2)

x = \sqrt{5.436*10^{-5}}

x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M

Dissociated form of  4-chlorobutanoic acid = C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M

Percentage dissociated = \frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100

Percentage dissociated = \frac{7.3729*10^{-3}}{1.8 }*100

Percentage dissociated = 0.4096

Percentage dissociated = 0.41%     (to two significant digits)

3 0
3 years ago
Read 2 more answers
Calculate the percent errorin a length measurementof 4.45cm if the correct value is 4.06
FinnZ [79.3K]

Answer:

9.6 %

Explanation:

<u>Step 1: How to define  percent error ? </u>

⇒ % error is the difference between a measured value and the known or accepted value

⇒Percent error is calculated using the following formula:

⇒%error  =  | Experimental value-theoretical value/theoretical value |  x100%

⇔ this can be written as well as : error = (| Experimental value/ theoretical value | - | Theoretical value / Theoretical value | ) x100%

<u>Step 2: Calculate % error</u>

In this case, this means :

%error = ( |(4.45 cm - 4.06cm ) / 4.06cm | ) x100%

%error = 0.096 x100%

%error =9.6 %

3 0
3 years ago
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