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allsm [11]
3 years ago
15

What is the formula for calcium carbonate?

Chemistry
1 answer:
mote1985 [20]3 years ago
6 0
The formula is CaCO3
You might be interested in
What is the change in density if a sample goes from 3.21 g/L to 5.43 g/mL?
Step2247 [10]

Answer:

\Delta \rho =2.22 g/mL

Explanation:

Hello,

In this case, since a change in science is widely known to be considered as a subtraction between the the final and initial values of two measured variables and is represented via Δ, here the final density is 5.43 g/mL and the initial one was 3.21 g/mL, therefore, the change in density is:

\Delta \rho=\rho _f-\rho _i\\ \\\Delta \rho=5.43g/mL-3.21g/mL\\\\\Delta \rho =2.22 g/mL

Best regards.

3 0
2 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
2 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
2 years ago
Mg(OH)2 + 2 HBr à MgBr2 + 2 H2O
AnnyKZ [126]

Explanation:

The balanced equation of the reaction is given as;

Mg(OH)2 (s) + 2 HBr (aq) → MgBr2 (aq) + 2 H2O (l)

1. How many grams of MgBr2 will be produced from 18.3 grams of HBr?

From the reaction;

2 mol of HBr produces 1 mol of  MgBr2

Converting to masses using;

Mass = Number of moles * Molar mass

Molar mass of HBr = 80.91 g/mol

Molar mass of MgBr2 = 184.113 g/mol

This means;

(2 * 80.91 = 161.82g) of HBr produces (1 * 184.113 = 184.113g) MgBr2

18.3g would produce x

161.82 = 184.113

18.3 = x

x = (184.113 * 18.3 ) / 161.82 = 20.8 g

2. How many moles of H2O will be produced from 18.3 grams of HBr?

Converting the mass to mol;

Number of moles = Mass / Molar mass = 18.3 / 80.91 = 0.226 mol

From the reaction;

2 mol of HBr produces 2 mol of H2O

0.226 mol would produce x

2 =2

0.226 = x

x = 0.226 * 2 / 2 = 0.226 mol

3. How many grams of Mg(OH)2 are needed to completely react with 18.3 grams of HBr?

From the reaction;

2 mol of HBr reacts with 1 mol of Mg(OH)2

18.3g of HBr =  0.226 mol

2 = 1

0.226 = x

x = 0.226 * 1 /2

x = 0.113 mol

5 0
2 years ago
True or false: the density of an object can be calculated by dividing its length by its width.
solong [7]
This is true!!!!!!!!!!!!!!!!!!!!!!!!
3 0
3 years ago
Read 2 more answers
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