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larisa86 [58]
3 years ago
12

If an object has a mass of 3500 mg what is the mass in kg? Show detailed work

Chemistry
1 answer:
dangina [55]3 years ago
7 0
The answer is 0.0035

In order to figure out kg, you have to know that 1 milligram is equal to 1/100000 kilograms. You take the amount of mg's that you have, which is 3500, and divide it by 1000000 which will give you your answer.

3500/1000000= 0.0035 kg
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A student measures the mass of a 6.0 cm3 block of brown sugar to be 10.0 g. What is the density of the brown sugar?
Alex17521 [72]

Answer:

1.67g/cm3

Explanation:

The formula for density is d=\frac{m}{v} . The m variable stands for mass and the v variable stands for volume.

The mass of the brown sugar is 10.0g and the volume is 6.0cm3, so we can plug those values into the equation.

d=\frac{10g}{6cm^{3} }

d=1.67\frac{g}{cm^{3} }

Rounded to 3 significant figures, the density of the block of brown sugar is 1.67 g/cm3. If the mass is in grams and the volume is in cm3, the unit for the final answer is \frac{g}{cm^{3} } (grams per centimetres cubed).

8 0
3 years ago
Read 2 more answers
S(s)+3F2(g)->SF6(g) how many mol of F2 are required to react completely with 2.30 mol of S?
Brilliant_brown [7]

Answer:  There are 6.9 mol of F_{2} are required to react completely with 2.30 mol of S.

Explanation:

The given reaction equation is as follows.

S(s) + 3F_{2}(g) \rightarrow SF_{6}(g)

Here, 1 mole of S is reaction with 3 moles of F_{2} which means 1 mole of S requires 3 moles of F_{2}.

Therefore, moles of F_{2} required to react completely with 2.30 moles S are calculated as follows.

1 mol S = 3 mol F_{2}\\2.30 mol S = 3 mol F_{2} \times 2.30 \\= 6.9 mol F_{2}

Thus, we can conclude that there are 6.9 mol of F_{2} are required to react completely with 2.30 mol of S.

3 0
3 years ago
Convert 5.28 x 1019 molecules of C6H1206 to grams.
nlexa [21]

Answer:

m=0.0158g

Explanation:

Hello there!

In this case, it is possible to comprehend these mass-particles problems by means of the concept of mole, molar mass and the Avogadro's number because one mole of any substance has 6.022x10²³ particles and have a mass equal to the molar mass.

In such a way, for C₆H₁₂O₆, whose molar mass is about 180.16 g/mol, the referred mass would be:

m=5.28x10^{19}molecules*\frac{1mol}{6.022x10^{23}molecules}*\frac{180.16g}{1mol}\\\\m=0.0158g

Best regards!

5 0
3 years ago
aspirin C6H4(CO2)(CO2CH3),can be prepared in the chemistry laboratory by the reactions of salicylic acid, C6H4(CO2H)(OH),with ac
Ket [755]
262mol 1=kg
g=1000
from stoichoimetry

x=102*1000/360
x=102000/360
x=283.33

density =m/v
=283.33/1.082
=262mol

8 0
3 years ago
There is a structure of CH3CH2CHCHCH2CH3, with a –CH2CH3 group attached to the third and fourth carbons.
NARA [144]

Explanation:

2,3-diethyl hexane

At first we select a long chain.

Then, we number that chain from that side where substituent position is closer.

Then, we write it's IUPAC name

Position of substituent + substituent name + chain name + suffix

Here,

2,3 + -diethyl + hex + -ane

= 2,3-diethyl hexane

7 0
2 years ago
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