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fredd [130]
3 years ago
5

Find the mass in 6.2 miles of lithium bromide, show work

Chemistry
1 answer:
Delicious77 [7]3 years ago
7 0

Answer:

if you meant moles this is the answer 1

mol of lithium bromide has a mass of

86.85g⋅mol−1. You have 2.6 moles

Explanation: mass = 86.85g⋅ mol−1× 2.6 mol

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A 56 g sample of peanut oil has a volume of 60.9 cm³. Calculate the density of the peanut oil. Remember to use the correct numbe
Ivan
0.92 because Mass divided by Volume = Density 
5 0
3 years ago
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A 5.00 L sample of air at 0 C is warmed to 100.0 C. What is the new volume of the air? First, identify V1.
Paladinen [302]

Answer : The new volume of the air is, 6.83 L

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

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

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=5.00L\\T_1=0^oC=(0+273)K=273K\\V_2=?\\T_2=100^oC=(100+273)K=373K

Putting values in above equation, we get:

\frac{5.00L}{273K}=\frac{V_2}{373K}\\\\V_2=6.83L

Therefore, the new volume of the air is, 6.83 L

6 0
3 years ago
How many atoms are in 1.00 mol of calcium
SVEN [57.7K]

Answer: 6.02214076 atoms Ca

Explanation:

Ca is monoatomic, so atoms in 1 mol = avogadro number

6 0
3 years ago
If 50 ml of 0.235 M NaCl solution is diluted to 200.0 ml what is the concentration of the diluted solution
Helen [10]

This is a straightforward dilution calculation that can be done using the equation

M_1V_1=M_2V_2

where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.

Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:

M_2=\frac{M_1V_1}{V_2}.

Substituting in our values, we get

\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].

So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.

5 0
3 years ago
A volume of 90.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2
Alja [10]

Answer:

0.823 M was the molarity of the KOH solution.

Explanation:

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To calculate the concentration of base , we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH.

We are given:

n_1=2\\M_1=1.50 M\\V_1=24.7 mL\\n_2=1\\M_2=?\\V_2=90.0 mL

Putting values in above equation, we get:

2\times \1.50 M\times 24.7 mL=1\times M_2\times 90.0 mL

M_2=\frac{2\times 1.50M\times 24.7 mL}{1\times 90.0 mL}=0.823 M

0.823 M was the molarity of the KOH solution.

7 0
3 years ago
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