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2 years ago

Which image shows 2-hexyne?

Chemistry

1 answer:

MaRussiya [10]2 years ago

4 0

Answer: C

Explanation:

A. Shows 3-Hexyne (NOT 2-HEXYNE)

B. Shows 7 carbons (too many) (NOT 2-HEXYNE)

C. Shows a triple bond (yne) and 6 carbons and it's on the second carbon (2-HEXYNE)

D. Shows two substitent on the second carbon but the triple bond is on the 3rd carbon so it's 2,2-dimethyl-3-heptyne (NOT 2-HEXYNE)

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A solid with a density of 3.57g/mL and mass of 19.5g is added to a graduated cylinder that contain 23.2mL of water. What is the

NNADVOKAT [17]

Answer:

Volume of liquid = 28.7 mL

Explanation:

Given data;

Density of solid = 3.57 g/ml

Mass of solid = 19.5 g

Volume of water = 23.2 mL

Total volume when solid is dropped into graduated cylinder= ?

Solution:

Density = mass/ volume

v = m/d

v = 19.5 g/ 3.57 g/ml

v = 5.5 mL

Volume of liquid = volume of water + volume of solid

Volume of liquid = 23.2 mL + 5.5 mL

Volume of liquid = 28.7 mL

3 0

3 years ago

In the image above, the girl is using her arm to lift a weight. Her arm is acting as a _______.

Vinvika [58]

B the girls arm will act as a pulley

6 0

3 years ago

how many moles of iodine should be added to 750 grams of carbon tetrachloride to prepare a 0.24 m solution

aleksley [76]

Answer:

0.18 mol

Explanation:

Given data

Mass of carbon tetrachloride (solvent): 750 g

Molality of the solution: 0.24 m

Moles of iodine (solute): ?

Step 1: Convert the mass of the solvent to kilograms

We will use the relationship 1 kg = 1,000 g.

$750g \times \frac{1kg}{1,000g} =0.750kg$

Step 2: Calculate the moles of the solute

The molality is equal to the moles of solute divided by the kilograms of solvent. Then,

$m = \frac{moles\ of\ solute }{kilograms\ of\ solvent} \\moles\ of\ solute = m \times kilograms\ of\ solvent = \frac{0.24mol}{kg} \times 0.750kg = 0.18 mol$

6 0

3 years ago

Read 2 more answers

A. Which reactant is the limiting reagent?

Tasya [4]

Answer:

a. Zinc is the limiting reactant.

b. $m_{ZnBr_2}^{by\ Zn}=162.61gZnBr_2$

c. $m_{Br_2}^{leftover}=6.6g$

Explanation:

Hello there!

a. In this case, when zinc metal reacts with bromine, the following chemical reaction takes place:

$Zn+Br_2\rightarrow ZnBr_2$

Thus, since zinc and bromine react in a 1:1 mole ratio, we can compute their reacting moles to identify the limiting reactant:

$n_{Zn}=47.2g*\frac{1mol}{65.38g} =0.722molZn\\\\n_{Br_2}=122g*\frac{1mol}{159.8g} =0.763molBr_2$

Thus, since zinc has the fewest moles we infer it is the limiting reactant.

b. Here, we compute the grams of zinc bromide via both reactants:

$m_{ZnBr_2}^{by\ Zn}=0.722molZn*\frac{1molZnBr_3}{1molZn} *\frac{225.22gZnBr_2}{1molZnBr_2} =162.61gZnBr_2\\\\m_{ZnBr_2}^{by\ Br_2}=0.763molBr_2*\frac{1molZnBr_3}{1molBr_2} *\frac{225.22gZnBr_2}{1molZnBr_2} =171.95gZnBr_2$