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tangare [24]
3 years ago
12

Can you measure tenths if a gram with confidence using balance?

Chemistry
1 answer:
hram777 [196]3 years ago
7 0
It depends on the calibration and sensitivity of the balance. If you’re using a lab balance, then yes. Most lab balances are calibrated to the hundredths at a minimum. Take time to tare the scale correctly, and give the sample enough time on the scale to reach a definite number.
You might be interested in
Complete the charge balance equation for an aqueous solution of h2co3 that ionizes to hco−3 and co2−3.
Zielflug [23.3K]

The charge balance equation for an aqueous solution of H₂CO₃ that ionizes to HCO₃⁻ and CO₃⁻² is [HCO₃⁻] =  2[CO₃⁻²] + [H⁺] + [OH⁻]

<h3>What is Balanced Chemical Equation ?</h3>

The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

The equation for aqueous solution of H₂CO₃ is

H₂CO₃ → H₂O + CO₂

The charge balance equation is

[HCO₃⁻] =  2[CO₃⁻²] + [H⁺] + [OH⁻]

Thus from the above conclusion we can say that The charge balance equation for an aqueous solution of H₂CO₃ that ionizes to HCO₃⁻ and CO₃⁻² is [HCO₃⁻] =  2[CO₃⁻²] + [H⁺] + [OH⁻]

Learn more about the Balanced Chemical equation here: brainly.com/question/26694427
#SPJ4

8 0
2 years ago
A 45-g aluminium spoon(specific heat 0.80 / J/gdegree Celsius) at 24 degree celsius is placed in 180 ml(180 grams) of coffee at
Vlad [161]

Explanation:

a) The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, heat_{absorbed}=heat_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 45 g

m_2 = mass of coffee = 180 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 85^oC

c_1 = specific heat of aluminium = 0.80J/g^oC

c_2 = specific heat of coffee= 4.186 J/g^oC

Putting all the values in equation 1, we get:

45 g\times 0.80J/g^oC\times (T_{final}-24^oC)=-[180 g\times 4.186J/g^oC\times (T_{final}-83^oC)]

T_{final}=80.30^oC

80.30 °C is the final temperature.

b) Energy flows from higher temperature to lower temperature.Whenever two bodies with different energies and temperature come in contact. And the resulting temperature of both bodies will less then the body with high temperature and will be more then the body with lower temperature.

So, is our final temperature of both aluminium and coffee that is 80°C less than initial temperature of coffee and more than the initial temperature of the aluminum.

8 0
3 years ago
If the starting volume of a hot air balloon is 55,500 m3and the initial temperature is 21 °C, what is the temperature inside the
Dmitry_Shevchenko [17]

Answer:

T₂ = 392 K

Explanation:

Given that,

Initial volume of the hot air balloon, V₁ = 55500 m³

Initial temperature, T₁ = 21°C = 294 K

Final volume, V₂ = 74000 m³

We need to find the final temperature inside the balloon. The relation between the temperature and volume is given by charles law i.e.

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

Where

T₂ is the final temperature

So,

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\T_2=\dfrac{T_1V_2}{V_1}\\\\T_2=\dfrac{294\times 74000 }{55500 }\\\\T_2=392\ K

So, the new temperature is 392 K.

8 0
3 years ago
How many moles of O2 are consumed if 20 moles of SO2 are produced
tatiyna

Answer:

25

I hope this helps

5 0
3 years ago
A student plans a two-step synthesis of 1-ethyl-3-nitrobenzene from benzene. The first step is nitration of benzene to give nitr
arlik [135]

Answer:

Nitrobenzene is too deactivated (by the nitro group) to undergo a Friedel-Crafts alkylation.

Explanation:

The benzene ring in itself does not easily undergo electrophilic substitution reaction. Some groups activate or deactivate the benzene ring towards electrophilic substitution reactions.

-NO2 ia a highly deactivating substituent therefore, Friedel-Crafts alkylation of nitrobenzene does not take place under any conditions.

This reaction scheme is therefore flawed because Nitrobenzene is too deactivated (by the nitro group) to undergo a Friedel-Crafts alkylation.

7 0
3 years ago
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