All categories

3 years ago

Rewrite 76.00 to have 1 sig fig.

Chemistry

1 answer:

Luba_88 [7]3 years ago

4 0

Answer:

76.00

sig fig:4

decmials:2

scientific notation:7.600 x 10 1

words: seventy-six

p.s that one is on the right top side of ten

Explanation:

You might be interested in

Ammonium phosphate ((NH4)3 PO4) is an important ingredient to many fertilizers. It can be made by reacting phosphoric acid (H3 P

lara31 [8.8K]

Answer:

15.35 g of (NH₄)₃PO₄

Explanation:

First we need to look at the chemical reaction:

3 NH₃ + H₃PO₄ → (NH₄)₃PO₄

Now we calculate the number of moles of ammonia (NH₃):

number of moles = mass / molecular wight

number of moles = 5.24 / 17 = 0.308 moles of NH₃

Now from the chemical reaction we devise the following reasoning:

if 3 moles of NH₃ are produce 1 mole of (NH₄)₃PO₄

then 0.308 moles of NH₃ are produce X moles of (NH₄)₃PO₄

X = (0.308 × 1) / 3 = 0.103 moles of (NH₄)₃PO₄

mass = number of moles × molecular wight

mass = 0.103 × 149 = 15.35 g of (NH₄)₃PO₄

4 0

3 years ago

If 65.5 moles of an ideal gas is at 9.15 atm at 50.30 °C, what is the volume of the gas?

qwelly [4]

To calculate for the volume, we need a relation to relate the number of moles (n), pressure (P), and temperature (T) with volume (V). For simplification, we assume the gas is an ideal gas. So, we use PV=nRT.

PV = nRT where R is the universal gas constant
V = nRT / P
V = 65.5 ( 0.08205 ) (273.15 + 50.30) / 9.15
V = 189.98 L

7 0

3 years ago

if you placed the contents of a packet of powdered iced tea mix into a bottle of water and shook it, which of the following woul

victus00 [196]

If you placed the contents of a packet of powdered iced tea mix into a bottle of water and shook it, D. The solution would be homogenous if half of the powdered solute sat at the bottom of the bottle.

3 0

3 years ago

Read 2 more answers

What is the molarity of a solution that is made by mixing 35.5 g of Ba(OH)2 in 325 ml of solution?

choli [55]

Answer:

$M=0.638M$

Explanation:

Hello!

In this case, since the molarity of a solution is calculated by diving the moles of solute by the volume of solution in liters, we first compute the moles of barium hydroxide in 35.5 g as shown below:

$n=35.5g Ba(OH)_2*\frac{1molBa(OH)_2}{171.34gBa(OH)_2}\\\\n=0.207mol$

Then, the liters of solution:

$V=325mL*\frac{1L}{1000mL} =0.325L$

Finally, the molarity turns out:

$M=\frac{0.207mol}{0.325L}\\\\M=0.638M$

Best regards!

5 0

3 years ago

Calculate the maximum volume (in mL) of 0.143 M HCl that each of the following antacid formulations would be expected to neutral

Nuetrik [128]

Answer:

a. The maximum volume of 0.143 M HCl required is 154.4 mL.

b. The maximum volume of 0.143 M HCl required is 135.7 mL.

Explanation:

$Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O$

Mass of aluminum hydroxide = 350 mg = 0.350 g ( 1mg = 0.001 g)

Moles of aluminum hydroxide = $\frac{0.350 g}{78 g/mol}=0.004487 mol$

According to reaction ,3 moles of HCl neutralize 1 mole of aluminum hydroxide.Then 0.004487 mole of aluminum hydroxide will be neutralize by :

$\frac{3}{1}\times 0.004487 mol=0.01346 mol$ of HCl.

$Mg(OH)_2+2HCl\rightarrow MgCL_2+2H_2O$

Mass of magnesium hydroxide = 250 mg = 0.250 g ( 1mg = 0.001 g)

Moles of magnesium hydroxide = $\frac{0.250 g}{58 g/mol}=0.004310 mol$

According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310 mole of magnesium hydroxide will be neutralize by :

$\frac{2}{1}\times 0.004310 mol=0.008621 mol$ of HCl.

Total moles of HCl required to neutralize both :

0.01346 mol + 0.008621 mol = 0.02208 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

$Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}$

$V=\frac{0.02208 mol}{0.143 M}=0.1544 L$

1 L = 1000 mL

0.1544 L = 154.4 mL

The maximum volume of 0.143 M HCl required is 154.4 mL.

$CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2$

Mass of calcium carbonate = 970mg = 0.970 g ( 1mg = 0.001 g)

Moles of calcium carbonate = $\frac{0.970 g}{100 g/mol}=0.00970 mol$

According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :

$\frac{2}{1}\times 0.00970 mol=0.0194 mol$ of HCl.

Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

$Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}$

$V=\frac{0.0194 mol}{0.143 M}=0.1357 L$

1 L = 1000 mL

0.1357 L = 135.7 mL

The maximum volume of 0.143 M HCl required is 135.7 mL.

4 0

3 years ago