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miskamm [114]
2 years ago
10

How would you find the mass of an object from its density and volume ?

Chemistry
1 answer:
Lina20 [59]2 years ago
7 0

Answer:

Mass= volume x density

Explanation:

Multiply the density by the volume.

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Hey, i need help. the teacher provided notes but they don’t explain ANYTHING.
Arlecino [84]

Answer:

I think

1. A

2. 400

3. 100

4. IDK srry

Explanation:

Ijust want to help, but I also want brainliest

5 0
3 years ago
1. A given mass of air has a volume of 6.00 L at 80.0°C. At constant pressure, the temperature is
earnstyle [38]

3 L will be the final volume for the gas as per Charle's law.

Answer:

Explanation:

The kinetic theory of gases has two significant law which forms the backdrop of motion of gases. They are Charle's law and Boyle's law. As per Charle's law, the volume of any gas molecule at constant pressure is directly proportional to the temperature of the molecule.

V∝ T

Since, here two volumes are given and at two different temperatures with constant pressure. Then as per Charle's law, the relation between the volumes of air at different temperature will be

\frac{V_{1} }{T_{1} }= \frac{V_{2} }{T_{2} }

So in this case, V1 = 6 L and T1 = 80° C. Similarly, T2 = 40° C. So we have to determine the V2.

\frac{6}{80}=\frac{V_{2} }{40}

V_{2}=\frac{6*40}{80}=3 L

So, 3 L will be the final volume for the gas as per Charle's law.

4 0
3 years ago
The range in size of most atomic radii is approximately _____.5 × 10−11 m to 2 × 10−10 m2 to 5 nm5 × 10−21 m to 2 × 10−20 m2 to
Komok [63]
A) 5*10-11 m to 2*10-10 m
4 0
3 years ago
A.) A student titrated a 15.00-mL sample of a solution containing a weak, monoprotic acid with NaOH. If the titration required 1
nikklg [1K]

Answer:

A) 0.1225 M

B) 100.4 g/mol

Explanation:

Step 1: Write the generic neutralization reaction

HA(aq) + NaOH(aq) ⇒ NaA(aq) + H₂O(l)

Step 2: Calculate the reacting moles of NaOH

17.73 mL of 0.1036 M NaOH react. The reacting moles are:

0.01773 L × 0.1036 mol/L = 1.837 × 10⁻³ mol

Step 3: Calculate the reacting moles of HA

The molar ratio of HA to NaOH is 1:1. The reacting moles of HA are 1/1 × 1.837 × 10⁻³ mol = 1.837 × 10⁻³ mol.

Step 4: Calculate the molar concentration of HA

1.837 × 10⁻³ moles of HA are in a 15.00 mL volume. The molar concentration is:

M = 1.837 × 10⁻³ mol / 0.01500 L = 0.1225 M

Step 5: Calculate the molar mass of HA

1.837 × 10⁻³ moles of HA weigh 0.1845 g. The molar mass of HA is:

0.1845 g / 1.837 × 10⁻³ mol = 100.4 g/mol

5 0
3 years ago
Hydrogen gas and nitrogen gas react to form ammonia gas. What volume of ammonia would be produced by this reaction if 5.53 mL of
lina2011 [118]

Answer:

5.53 mL of hydrogen will produce 3.6867 mL of Ammonia

Explanation:

The complete balance equation for the given reaction is

N2 + 3H2 --> 2 NH3

Thus, 3 moles of hydrogen produces 2 moles of NH3

Hence, the volume of ammonia produced = 5.53 * (2/3) = 3.6867 mL

Hence, 5.53 mL of hydrogen will produce 3.6867 mL of Ammonia

6 0
3 years ago
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