Table salt is 42.7% sodium how many grams of salt contain 76 g of sodium

If the same amount of heat is added to 25.0 g of each of the metals, which are all at the same initial temperature, which metal

AVprozaik [17]

Answer:

The bismuth sample.

Explanation:

The specific heat $c$ of a substance (might not be a metal) is the amount of heat required for heating a unit mass of this substance by unit temperature (e.g., $\rm 1\; ^{\circ}C$ .) The formula for specific heat is:

$\displaystyle c = \frac{Q}{m \cdot \Delta T}$ ,

where

$Q$ is the amount of heat supplied.
$m$ is the mass of the sample.
$\Delta T$ is the increase in temperature.

In this question, the value of $Q$ (amount of heat supplied to the metal) and $m$ (mass of the metal sample) are the same for all four metals. To find $\Delta T$ (change in temperature,) rearrange the equation:

$\displaystyle c \cdot \Delta T = \frac{Q}{m}$ ,

$\displaystyle \Delta T = \frac{Q}{c \cdot m}$ .

In other words, the change in temperature of the sample, $\Delta T$ can be expressed as a fraction. Additionally, the specific heat of sample, $c$ , is in the denominator of that fraction. Hence, the value of the fraction would be the largest for sample with the smallest specific heat.

Make sure that all the specific heat values are in the same unit. Find the one with the smallest specific heat: bismuth ( $\rm 0.123 \; J \cdot g\cdot \,^{\circ}C^{-1}$ .) That sample would have the greatest increase in temperature. Since all six samples started at the same temperature, the bismuth sample would also have the highest final temperature.

3 0

3 years ago

How are the African plate and south American plate moving relative to each other

Kamila [148]

They are moving away from each other

6 0

3 years ago

1.20×10−8s to nanoseconds

AveGali [126]

Answer:

There are 12 nanoseconds in $1.2\times 10^{-8}\ s$ .

Explanation:

We need to convert $1.2\times 10^{-8}\ s$ to nanoseconds.

We know that,

$1\ s=10^9\ ns$

Now using unitary method to solve it such that,

$1.2\times 10^{-8}\ s=1.2\times 10^{-8}\ \times 10^9\\\\=1.2\times 10\\\\=12\ ns$

So, there are 12 nanoseconds in $1.2\times 10^{-8}\ s$ .

7 0

3 years ago

Calculate either [ H 3 O + ] or [ OH − ] for each of the solutions.

STALIN [3.7K]

Answer: Solution A : $[H_3O^+]=0.300\times 10^{-7}M$

Solution B : $[OH^-]=0.107\times 10^{-5}M$

Solution C : $[OH^-]=0.177\times 10^{-10}M$

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.

$pH=-\log[H_3O^+]$

$pOH=-log[OH^-}$

$pH+pOH=14$

$[H_3O^+][OH^-]=10^{-14}$

a. Solution A: $[OH^-]=3.33\times 10^{-7}M$

$[H_3O^+]=\frac{10^{-14}}{3.33\times 10^{-7}}=0.300\times 10^{-7}M$

b. Solution B : $[H_3O^+]=9.33\times 10^{-9}M$

$[OH^-]=\frac{10^{-14}}{9.33\times 10^{-9}}=0.107\times 10^{-5}M$

c. Solution C : $[H_3O^+]=5.65\times 10^{-4}M$

$[OH^-]=\frac{10^{-14}}{5.65\times 10^{-4}}=0.177\times 10^{-10}M$

7 0

3 years ago

If 10.0 mL of a .600 M of HNO3 reacts with 31.0 mL of .700M Ba(OH)2 solution, what is the molarity of Ba(OH)2 after the reaction

Tasya [4]

Answer:

0.456M

Explanation:

1. Balanced molecular equation

$2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_2+2H_2O$

2. Mole ratio

$\dfrac{2molHNO_3}{1molBa(OH)_2}$

3. Moles of HNO₃

Number of moles = Molarity × Volume in liters
n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃

4. Moles Ba(OH)₂

n = 0.700M × 0.0310 liter = 0.0217 mol

5. Limiting reactant

Actual ratio:

$\dfrac{0.0600molHNO_3}{0.0217molBa(OH)_2}\approx0.28$

Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.

Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.

6. Final molarity of Ba(OH)₂

Molarity = number of moles / volume in liters
Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M

5 0

3 years ago