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tigry1 [53]
2 years ago
7

I’ll give 5 stars and Brainliest plus this is 20 points lol

Chemistry
1 answer:
Daniel [21]2 years ago
4 0

Answer:

I think is b

Explanation:

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Mary starts from her house, walks 80 meters south, and stops to chat with her aunt on the sidewalk. After chatting for a few min
Olenka [21]
(80+125+45) / 10 = 250/10 =25
25 meters per minute= 0.41 meters/second

the direction and stopping time is irrelevant to the problem.
6 0
3 years ago
Use the information below, to write a word equation, then change it into a balanced symbol equation.
Kamila [148]

Answer:

Mg    +  2HCl  -> MgCl2   + H2

Explanation:

7 0
3 years ago
how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate
Stolb23 [73]

Answer:- 0.273 kg

Solution:- A double replacement reaction takes place. The balanced equation is:

3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})

= 95.7gNaClO_3

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate  and then the grams are converted to kg.

95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

7 0
3 years ago
One litre of hydrogen at STP weight 0.09gm of 2 litre of gas at STP weight 2.880gm. Calculate the vapour density and molecular w
expeople1 [14]

Answer:

we know, at STP ( standard temperature and pressure).

we know, volume of 1 mole of gas = 22.4L

weight of 1 Litre of hydrogen gas = 0.09g

so, weight of 22.4 litres of hydrogen gas = 22.4 × 0.09 = 2.016g ≈ 2g = molecular weight of hydrogen gas.

similarly,

weight of 2L of a gas = 2.88gm

so, weight of 22.4 L of the gas = 2.88 × 22.4/2 = 2.88 × 11.2 = 32.256g

hence, molecular weight of the gas = 32.256g

vapor density = molecular weight/2

= 32.256/2 = 16.128g

hence, vapor density of the gas is 16.128g.

Explanation:

4 0
2 years ago
Object A has a mass of 500g and object B has a mass of 300g. Which object is more dense?
KonstantinChe [14]

Answer:

500g because 300g is less dense than 500g its kinda like a weight

Explanation:

i

4 0
3 years ago
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