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nikklg [1K]
2 years ago
10

How many atm are in 1520mmHg​

Chemistry
1 answer:
stealth61 [152]2 years ago
8 0

Answer:

2

Explanation:

millimetre of mercury =1520

divide the pressure by 760 (do this always)

1520/760 = 2

Thats how you do it!

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What sample size (grams) of Na3PO4 (FW 164.00) known to be 50.00% pure should be used to consume exactly 40.00 mL of 0.1000 M HC
Mkey [24]

Answer:

0.109 g.

Explanation:

Equation of the reaction:

Na3PO4 + 3HCl --> 3NaCl + H3PO4

Number of moles of HCl = molar concentration × volume

= 0.1 × 0.04

= 0.004 mol.

By stoichiometry, 1 mole of Na3PO4 neutralises 3 moles of HCl. Therefore, number of moles of Na3PO4 = 0.004/3

= 0.0013 mol

Mass of Na3PO4 = molar mass × number of moles

= 0.0013 × 164

= 0.219 g

Since 50% of Na3PO4 was present in the sample. Let 100 g be the total mass of the substance

= 0.219 × 50 g/100 g

= 0.109 g.

3 0
3 years ago
In Balancing Equations, why do I put a 3 before reactant H?
Katena32 [7]
You have 3 (h2(so4)) on the reactants side so you need to have 6 total hydrogen’s on the products side. Therefore 3(h2) is required.
5 0
3 years ago
A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling
aniked [119]

<u>Answer:</u> The mass of glucose in final solution is 0.420 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}        .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated glucose solution

M_2\text{ and }V_2 are the molarity and volume of diluted glucose solution

We are given:

M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL

Putting values in above equation, we get:

0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g

Hence, the mass of glucose in final solution is 0.420 grams

3 0
3 years ago
9. What is the name of the molecule?
AnnyKZ [126]

Answer:

3–methyl–2–butanol

Explanation:

To name the compound, we must:

1. Identify the functional group.

2. Give the functional group of the compound the lowest possible count.

3. Locate the longest continuous carbon chain. This gives the parent name of the compound.

4. Identify the substituent group attached.

5. Give the substituent group the lowest possible count.

6. Combine the above to get the name of the compound.

Now, let us obtain the name of the compound.

1. The functional group of the compound is Alcohol i.e —OH.

2. The functional group is located at carbon 2.

3. The longest continuous carbon chain is carbon 4 i.e butane. But the presence of the functional group i.e OH will replace the –e in butane with –ol. Therefore, the compound is butanol.

4. The substituent group attached is methyl i.e CH3.

5. The substituent group is located at carbon 3.

6. Therefore, the name of the compound is:

3–methyl–2–butanol.

8 0
3 years ago
PLEASE HELP WILL GIVE BRAINLEST
aliina [53]

Answer:

Option A.

Lower air pressure results in a lower boiling point

Explanation:

This is because in an open system, the lower the pressure the lesser the energy that will be required for boiling point. The is little or no collision of air molecules with the surface of the liquid

But if there is increase in pressure, more energy will be required to get to boiling point because there will be strong collision between air molecules and surface of the liquid.

8 0
3 years ago
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