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nikdorinn [45]
3 years ago
11

What lab equipment do you use to measure the density of gold and how to use it​

Chemistry
1 answer:
tigry1 [53]3 years ago
8 0

Answer: Divide the mass of the object by its volume to yield the density of the object. For example, the object density = mass / (Vf – Vi). Compare the measured density to that of pure gold (19.3 g/cc) to determine whether the object is made of pure gold.

Explanation: I searched it

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In water, Vanillin, C8H8O3, has a solubility of 0.070 moles of vanillin per liter of solution at 25C. What will be produced if 5
Rufina [12.5K]

Answer:

The full amount (5.00 g) will be dissolved in 1 L of water at 25°C.

Explanation:

The molecular weight (MW) of Vanillin (C₈H₈O₃) is calculated from the chemical formula as follows:

MW(C₈H₈O₃) = (12 g/mol x 8) + (1 g/mol x 8) + (16 g/mol x 3) = 152 g/mol

If 0.070 mol of C₈H₈O₃ are soluble per liter of water at 25°C, the maximum mass that can be dissolved in 1 L is:

0.070 mol x 152 g/mol = 10.64 g

Since 5.00 g is lesser than the maximum amount that can be dissolved (10.64 g), the added amount will be completely dissolved in 1 L of water at 25°C.

7 0
3 years ago
PLSSS HELP ANYONE ASAP!
algol [13]
B I hope it’s right I don’t really help a lot but yeah lol
7 0
3 years ago
The decomposition of ethyl amine, C2H5NH2, occurs according to the reaction: C2H5NH2(g)⟶C2H4(g)+NH3(g) At 85∘C, the rate constan
statuscvo [17]

Answer:

2.772 seconds

Explanation:

Given that;

t1/2 = 0.693/k

Where;

t1/2 = half life of the reaction

k= rate constant

Note that decomposition is a first order reaction since the rate of reaction depends on the concentration of one reactant

t1/2 = 0.693/2.5 x 10-1 s-1

t1/2= 2.772 seconds

7 0
3 years ago
Determine the mass of CaCO3 required to produce 40.0 mL CO2 at STP. Hint use molar volume of an ideal gas (22.4 L)
cupoosta [38]

Answer:

m_{CaCO_3}=0.179gCaCO_3

Explanation:

Hello,

In this case, since the undergoing chemical reaction is:

CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:

PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2

Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:

m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3

Regards.

3 0
4 years ago
Crude oil may contain hundreds of different types of hydrocarbons. Some examples include: Butane (C₄H₁₀)
miv72 [106K]

Answer:

The answer to your question is:

Explanation:

Butane (C₄H₁₀)                   Compound      It's composed by C and H

Dodecane (C₁₂H₂₆)            Compond        It's composed by C and H

Octane (C₈H₁₈)                   Compound      It's composed by C and H

Benzene (C₆H₆)                Compound      It's composed by C and H

Oxygen (O₂)                     Element           It's only oxygen

Carbon dioxide (CO₂)      Compound      It's composed by C and O

Water (H₂O)                      Compound      It's composed by H and O

Gasoline                           Mixture             It's composed by many hydrocarbons

Kerosene                          Mixture            It's a mixture of hydrocarbons

4 0
3 years ago
Read 2 more answers
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