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4 years ago

6

What unit do we use to measure work? A Grams B joules C Newton’s

1 answer:

vladimir1956 [14]4 years ago

4 0

Answer:

B

Explanation:

You might be interested in

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

Which of the following would be one of the interacting parts in a weather system?

zhenek [66]

CLOUDS would be one of the interacting parts in a weather system.

6 0

3 years ago

Write the net ionic equation for the precipitation reaction that occurs when aqueous solutions of chromium(III) bromide and pota

expeople1 [14]

Explanation:

An ionic equation will be the one in which all the participating species will be present as ions.

The given reaction will be as follows.

$CrBr_{3} + K_{2}CO_{3} \rightarrow Cr_{2}(CO_{3})_{3}(s) + KBr$

Balancing this equation by multiplying $CrBr_{3}$ by 2 and $K_{2}CO_{3}$ by 3 on reactant side. Whereas multiply KBr by 6 on product side.

$2CrBr_{3} + 3K_{2}CO_{3} \rightarrow Cr_{2}(CO_{3})_{3}(s) + 6KBr$

Hence, the net ionic equation will be as follows.

$2Cr^{3+}(aq) + 3CO^{2-}_{3}(aq) + 6K^{+} + 6Br^{-} \rightarrow Cr_{2}(CO_{3})_{3}(s) + 6K^{+} + 6Br^{-}$

As both $K^{+}$ and $Br^{-}$ are spectator ions. Hence, the net ionic equation will be as follows.

$2Cr^{3+}(aq) + 3CO^{2-}_{3}(aq) \rightarrow Cr_{2}(CO_{3})_{3}(s)$

4 0

3 years ago

the formula for sodium hydroxide is NaOH; the formula for potassium hydroxide is KOH. which two elements do you think are contai

Luden [163]

Answer:

oxygen and hydrogen :)

Explanation:

4 0

3 years ago

How many moles at STP would take up 44.8 liters?

Tom [10]

Answer:

n = 2 mol

Explanation:

Given data:

Pressure = standard = 1 atm

Temperature = standard = 273.15 K

Volume = 44.8 L

Number of moles = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

1 atm × 44.8 L = n × 0.0821 atm.L/ mol.K × 273.15 K

44.8 atm.L = n × 22.43 atm.L/ mol

n = 44.8 atm.L / 22.43 atm.L/ mol

n = 2 mol

4 0

3 years ago